Class 10: Arithmetic Progressions

$\displaystyle \textbf{Question 1. }\text{Cable cars at hill stations are one of the major tourist attractions.}$
$\displaystyle \text{On a hill station, the length of cable car ride from base point to top most point}$
$\displaystyle \text{on the hill is }5000\text{ m. Poles are installed at equal intervals on the way to provide}$
$\displaystyle \text{support to the cables on which car moves. The distance of first pole from base point}$
$\displaystyle \text{is }200\text{ m and subsequent poles are installed at equal interval of }150\text{ m. Further,}$
$\displaystyle \text{the distance of last pole from the top is }300\text{ m. Based on the above information,}$
$\displaystyle \text{answer the following questions using Arithmetic Progression:} \hspace{0.2cm}\text{[CBSE 2025]}$
$\displaystyle \text{(i) Find the distance of }10^{\mathrm{th}}\text{ pole from the base.}$
$\displaystyle \text{(ii) Find the distance between }15^{\mathrm{th}}\text{ pole and }25^{\mathrm{th}}\text{ pole.}$
$\displaystyle \text{(iii) (a) Find the time taken by cable car to reach }15^{\mathrm{th}}\text{ pole from the top if it is }$
$\displaystyle \text{moving at the speed of }5\text{ m/sec and coming from top.}$
$\displaystyle \text{OR}$
$\displaystyle \text{(iii) (b) Find the total number of poles installed along the entire journey.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Here,}$
$\displaystyle a_1=200,\ a_2=350,\ a_3=500$
$\displaystyle d=a_2-a_1=150,\ S_n=5000$
$\displaystyle \text{(i)}$
$\displaystyle a_{10}=a+9d=200+9\times150$
$\displaystyle =200+1350=1550\text{ m}$
$\displaystyle \text{(ii)}$
$\displaystyle a_{25}-a_{15}=a+24d-a-14d$
$\displaystyle =10\times150=1500\text{ m}$
$\displaystyle \text{(iii) (a) We have to find time taken by cable car to}$
$\displaystyle \text{reach }15^{\text{th}}\text{ pole from the top if speed }=5\text{ m/sec.}$
$\displaystyle a_{15}\text{ from top}=300+14\times150$
$\displaystyle =300+2100=2400\text{ m}$
$\displaystyle \text{As}\quad \text{Speed}=\frac{\text{Distance travelled}}{\text{Time taken}}$
$\displaystyle 5=\frac{2400}{t}$
$\displaystyle \Rightarrow t=\frac{2400}{5}=480\text{ sec}$
$\displaystyle \textbf{OR}$
$\displaystyle \text{(iii) (b) Total no of poles installed }=\frac{5000-500}{n}+1$
$\displaystyle \text{when }n\text{ is gap}$
$\displaystyle =\frac{4500}{150}+1$
$\displaystyle =30+1$
$\displaystyle =31\text{ poles.}$
$\\$

$\displaystyle \textbf{Question 2. }\text{A school has decided to plant some endangered trees on }51^{\mathrm{st}}$
$\displaystyle \text{World Environment Day in the nearest park. They have decided to plant those trees}$
$\displaystyle \text{in few concentric circular rows such that each succeeding row has }20\text{ more trees}$
$\displaystyle \text{than the previous one. The first circular row has }50\text{ trees. Based on the above}$
$\displaystyle \text{given information, answer the following questions:} \hspace{0.2cm}\text{[CBSE 2024(C)]}$
$\displaystyle \text{(i) How many trees will be planted in the }10^{\mathrm{th}}\text{ row?}$
$\displaystyle \text{(ii) How many more trees will be planted in the }8^{\mathrm{th}}\text{ row than in the }5^{\mathrm{th}}\text{ row?}$
$\displaystyle \text{(iii) (a) If }3200\text{ trees are to be planted in the park, then how many rows}$
$\displaystyle \text{are required?}$
$\displaystyle \text{OR}$
$\displaystyle \text{(iii) (b) If }3200\text{ trees are to be planted in the park, then how many trees}$
$\displaystyle \text{are still left to be planted after the }11^{\mathrm{th}}\text{ row?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Here }a=50\text{ and }d=20$
$\displaystyle \text{(i) Number of trees planted in }10^{\text{th}}\text{ row}$
$\displaystyle =a_{10}=50+9\times20=230$
$\displaystyle \text{(ii) }a_8-a_5=3\times20=60$
$\displaystyle \text{(iii) (a) Let }S_n=3200$
$\displaystyle \Rightarrow \frac{n}{2}[2\times50+(n-1)\times20]=3200$
$\displaystyle \Rightarrow n^2+4n-320=0$
$\displaystyle \Rightarrow (n+20)(n-16)=0$
$\displaystyle \Rightarrow n\neq-20$
$\displaystyle \Rightarrow n=16$
$\displaystyle \text{Hence, required number of rows are }16.$
$\displaystyle \textbf{OR}$
$\displaystyle \text{(iii) (b) Required number of trees left }=S_n-S_{11}$
$\displaystyle =3200-\frac{11}{2}[2\times50+10\times20]$
$\displaystyle =1550$
$\displaystyle \text{Hence, number of trees left are }1550.$
$\\$

$\displaystyle \textbf{Question 3. }\text{If }k+7,2k-2\text{ and }2k+6\text{ are three consecutive terms of an A.P.,} \\ \text{then the value of }k\text{ is:} \hspace{0.2cm}\text{[CBSE 2024(C)]}$
$\displaystyle \text{(a) }15 \qquad \text{(b) }17 \qquad \text{(c) }5 \qquad \text{(d) }1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) }\because k+7,\ 2k-2\text{ and }2k+6\text{ are in A.P.}$
$\displaystyle \therefore (2k-2)-(k+7)=(2k+6)-(2k-2)$
$\displaystyle \Rightarrow k-9=8$
$\displaystyle \Rightarrow k=17$
$\\$

$\displaystyle \textbf{Question 4. }\text{The }7^{\mathrm{th}}\text{ term from the end of the A.P.: }-8,-5,-2,\ldots,49\text{ is:} \hspace{0.2cm}\text{[CBSE 2024(C)]}$
$\displaystyle \text{(a) }67 \qquad \text{(b) }13 \qquad \text{(c) }31 \qquad \text{(d) }10$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(c) For 7th term from end}$
$\displaystyle a=49,\ d=-3$
$\displaystyle \therefore \text{7th term from end}=49+(7-1)(-3)$
$\displaystyle =49-18=31$
$\\$

$\displaystyle \textbf{Question 5. }\text{In an AP, if the first term }a=7,\ n\text{th term }a_{n}=84\text{ and the sum of first } \\ n\text{ terms }S_{n}=\frac{2093}{2},\text{ then }n\text{ is equal to:} \hspace{0.2cm}\text{[CBSE 2024]}$
$\displaystyle \text{(a) }22 \qquad \text{(b) }24 \qquad \text{(c) }23 \qquad \text{(d) }26$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(c) We know}$
$\displaystyle S_n=\frac{n}{2}(a+a_n)$
$\displaystyle \Rightarrow \frac{2093}{2}=\frac{n}{2}(7+84)$
$\displaystyle \Rightarrow 2093=91n$
$\displaystyle \Rightarrow n=23$
$\\$

$\displaystyle \textbf{Question 6. }\text{The sum of first and eighth terms of an AP is }32\text{ and their product is }60.$
$\displaystyle \text{Find the first term and common difference of the AP. Hence, also find the sum of its}$
$\displaystyle \text{first }20\text{ terms.} \hspace{0.2cm}\text{[CBSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be the first term and }d\text{ be the common difference of the given A.P.}$
$\displaystyle \text{A.T.Q. }a+a_8=32$
$\displaystyle \Rightarrow a+a+7d=32\quad [\because a_n=a+(n-1)d]$
$\displaystyle \Rightarrow 2a+7d=32 \qquad (i)$
$\displaystyle \text{Also, }a\times a_8=60$
$\displaystyle \Rightarrow a(a+7d)=60$
$\displaystyle \Rightarrow a^2+7ad=60 \qquad (ii)$
$\displaystyle \text{Now, from }(i),\text{ we get}$
$\displaystyle d=\frac{32-2a}{7} \qquad (iii)$
$\displaystyle \text{Substituting the value of }d\text{ from }(iii)\text{ in }(ii),\text{ we get}$
$\displaystyle a^2+7a\left(\frac{32-2a}{7}\right)=60$
$\displaystyle \Rightarrow a^2+32a-2a^2=60$
$\displaystyle \Rightarrow a^2-32a+60=0$
$\displaystyle \Rightarrow (a-30)(a-2)=0$
$\displaystyle \Rightarrow a=30\text{ or }a=2$
$\displaystyle \text{When }a=30,\text{ then }d=\frac{32-2\times30}{7}=-4$
$\displaystyle \text{When }a=2,\text{ then }d=\frac{32-2\times2}{7}=4$
$\displaystyle \text{So, first term}=30,\text{ common difference}=-4\text{ or first term}=2,\text{ common difference}=4$
$\displaystyle \text{Sum of first }n\text{ terms of an A.P. is given by,}$
$\displaystyle S_n=\frac{n}{2}[2a+(n-1)d]$
$\displaystyle \text{Taking }a=30\text{ and }d=-4$
$\displaystyle S_{20}=\frac{20}{2}[2\times30+19\times(-4)]=-160$
$\displaystyle \text{Taking }a=2\text{ and }d=4$
$\displaystyle S_{20}=\frac{20}{2}[2\times2+19\times4]=800$
$\\$

$\displaystyle \textbf{Question 7. }\text{In an AP of }40\text{ terms, the sum of first }9\text{ terms is }153\text{ and the sum}$
$\displaystyle \text{of last }6\text{ terms is }687. \text{Determine the first term and common difference of AP.}$
$\displaystyle \text{Also find the sum of all the terms of the AP.} \hspace{0.2cm}\text{[CBSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be the first term and }d\text{ be the common difference of the given A.P.}$
$\displaystyle \text{Sum of first }n\text{ terms of an A.P. is given by,}$
$\displaystyle S_n=\frac{n}{2}[2a+(n-1)d]$
$\displaystyle \text{Now, }S_9=153$
$\displaystyle \Rightarrow \frac{9}{2}(2a+8d)=153$
$\displaystyle \Rightarrow a+4d=17 \qquad (i)$
$\displaystyle \text{Last }6\text{ terms be }35^{\text{th}},36^{\text{th}},\ldots,40^{\text{th}}\text{ terms i.e. }a+34d,a+35d,a+36d,a+37d,a+38d,a+39d$
$\displaystyle \text{Then }(a+34d)+(a+35d)+(a+36d)+(a+37d)+(a+38d)+(a+39d)=687$
$\displaystyle \Rightarrow 6a+219d=687$
$\displaystyle \Rightarrow 2a+73d=229 \qquad (ii)$
$\displaystyle \text{Multiplying }(i)\text{ by }2,\text{ we get}$
$\displaystyle 2a+8d=34 \qquad (iii)$
$\displaystyle \text{Subtracting }(iii)\text{ from }(ii),\text{ we get}$
$\displaystyle 65d=195\Rightarrow d=3$
$\displaystyle \text{Putting }d=3\text{ in }(i),\text{ we get}$
$\displaystyle a+12=17\Rightarrow a=5$
$\displaystyle \therefore \text{First term, }a=5\text{ and common difference, }d=3$
$\displaystyle \text{Now, }S_{40}=\frac{40}{2}(2\times5+39\times3)$
$\displaystyle =20\times127$
$\displaystyle =2540$
$\\$

$\displaystyle \textbf{Questionv8. }\text{The next term of the A.P.: }\sqrt{6},\sqrt{24},\sqrt{54}\text{ is:} \hspace{0.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{(a) }\sqrt{60} \qquad \text{(b) }\sqrt{96} \qquad \text{(c) }\sqrt{72} \qquad \text{(d) }\sqrt{216}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) }\sqrt{6},\ \sqrt{24},\ \sqrt{54}$
$\displaystyle \text{We can write }\sqrt{6},\ 2\sqrt{6},\ 3\sqrt{6}$
$\displaystyle \therefore \text{Next term is }4\sqrt{6}=\sqrt{96}$
$\\$

$\displaystyle \textbf{Question 9. }\text{If }x+1,3x\text{ and }4x+2\text{ are three consecutive terms of an A.P., then} \\ \text{the value of }x\text{ is:} \hspace{0.2cm}\text{[CBSE 2023(C)]}$
$\displaystyle \text{(a) }2 \qquad \text{(b) }3 \qquad \text{(c) }4 \qquad \text{(d) }5$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) Let }t_{1}=x+1,\ t_{2}=3x\text{ and }t_{3}=4x+2$
$\displaystyle \text{Since, }t_{1},t_{2}\text{ and }t_{3}\text{ are in A.P.}$
$\displaystyle \therefore t_{2}-t_{1}=t_{3}-t_{2}$
$\displaystyle \Rightarrow 3x-x-1=4x+2-3x$
$\displaystyle \Rightarrow 2x-1=x+2$
$\displaystyle \Rightarrow x=3$
$\\$

$\displaystyle \textbf{Question 10. }\text{How many terms are there in an A.P. whose first and fifth terms are }$
$\displaystyle -14\text{ and }2, \text{ respectively and the }\text{last term is }62? \hspace{0.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{1st term of the A.P.}=-14$
$\displaystyle \text{and common difference}=d$
$\displaystyle \text{A.T.Q.}$
$\displaystyle a_5=2$
$\displaystyle \Rightarrow a+4d=2\Rightarrow -14+4d=2$
$\displaystyle \Rightarrow d=4$
$\displaystyle \text{Let there are }n\text{ terms in the A.P.}$
$\displaystyle \text{Last term}=62$
$\displaystyle a_n=62\Rightarrow a+(n-1)d=62$
$\displaystyle \Rightarrow -14+(n-1)(4)=62$
$\displaystyle \Rightarrow (n-1)=\frac{76}{4}=19$
$\displaystyle \Rightarrow n=20$
$\\$

$\displaystyle \textbf{Question 11. }\text{Which term of the A.P.: }65,61,57,53,\ldots\text{ is the first negative term?} \\ \hspace{0.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given A.P. is }65,61,57,53,\ldots$
$\displaystyle \text{Here, }a=65,\ d=61-65=-4$
$\displaystyle \text{Let }a_n\text{ is 1st negative term.}$
$\displaystyle a_n<0$
$\displaystyle \Rightarrow a+(n-1)d<0$
$\displaystyle \Rightarrow 65+(n-1)(-4)<0$
$\displaystyle \Rightarrow (n-1)(-4)<-65$
$\displaystyle \Rightarrow (n-1)>\frac{65}{4}$
$\displaystyle \Rightarrow n>\frac{65}{4}+1=\frac{69}{4}$
$\displaystyle \Rightarrow n>17\frac{1}{4}$
$\displaystyle \therefore \text{1st negative term is 18th term.}$
$\\$

$\displaystyle \textbf{Question 12. }\text{Find the sum of all integers between }50\text{ and }500,\text{ which} \\ \text{are divisible by }7. \hspace{0.2cm}\text{[CBSE 2023(C)]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The list of integers between }50\text{ and }500\text{ that are divisible by }7:$
$\displaystyle 56,63,70,\ldots,497$
$\displaystyle \text{The above list forms an A.P. with,}$
$\displaystyle \text{First term, }a=56$
$\displaystyle \text{Common difference, }d=7$
$\displaystyle \text{Now, }a_n=497$
$\displaystyle a+(n-1)d=497$
$\displaystyle \Rightarrow 56+(n-1)\times7=497$
$\displaystyle \Rightarrow n=64$
$\displaystyle \text{Now,}$
$\displaystyle S_n=\frac{n}{2}[2a+(n-1)d]$
$\displaystyle \therefore S_{64}=\frac{64}{2}[2\times56+63\times7]$
$\displaystyle =17696$
$\\$

$\displaystyle \textbf{Question 13. }\text{How many numbers lie between }10\text{ and }300,\text{ which when divided by }$
$\displaystyle 4\text{ leave a remainder }3?\text{ Also, find } \text{their sum.} \hspace{0.2cm}\text{[CBSE 2023(C)]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The list of numbers that lie between }10\text{ and }300\text{ and leave a remainder }3\text{ when divided by }4\text{ is}$
$\displaystyle 11,15,19,\ldots,299$
$\displaystyle \text{The above list is an A.P. with}$
$\displaystyle \text{first term, }a=11\text{ and common difference, }d=4$
$\displaystyle \text{Now, }a_n=299$
$\displaystyle a+(n-1)d=299$
$\displaystyle \Rightarrow 11+(n-1)\times4=299$
$\displaystyle \Rightarrow n=73$
$\displaystyle \text{Also,}$
$\displaystyle S_n=\frac{n}{2}[2a+(n-1)d]$
$\displaystyle \therefore S_{73}=\frac{73}{2}[2\times11+72\times4]$
$\displaystyle =11315$
$\\$

$\displaystyle \textbf{Question 14. }\text{Assertion (A): }a,b,c\text{ are in A.P. if and only if }2b=a+c. \hspace{0.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{Reason (R): The sum of first }n\text{ odd natural numbers is }n^2.$
$\displaystyle \text{(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).}$
$\displaystyle \text{(b) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).}$
$\displaystyle \text{(c) Assertion (A) is true but Reason (R) is false.}$
$\displaystyle \text{(d) Assertion (A) is false but Reason (R) is true.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For three numbers }a,b,c\text{ to be in A.P., the middle term must be the arithmetic mean of the other two.}$
$\displaystyle \therefore b=\frac{a+c}{2}$
$\displaystyle \therefore 2b=a+c$
$\displaystyle \therefore \text{Assertion (A) is true.}$
$\displaystyle \text{Also, the sum of the first }n\text{ odd natural numbers is }n^2.$
$\displaystyle \therefore \text{Reason (R) is true.}$
$\displaystyle \text{However, Reason (R) does not explain Assertion (A).}$
$\displaystyle \therefore \text{Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).}$
$\displaystyle \therefore \text{Correct option is (b).}$
$\\$

$\displaystyle \textbf{Question 15. }\text{Which term of the AP }-\frac{11}{2},-3,-\frac{1}{2},\ldots\text{ is }\frac{49}{2}? \hspace{0.2cm}\text{[CBSE 2022]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let nth term of given A.P. is }\frac{49}{2}$
$\displaystyle \text{Given A.P. is }-\frac{11}{2},\ -3,\ -\frac{1}{2}\ldots$
$\displaystyle \text{Here }a=-\frac{11}{2},\ d=-3-\left(-\frac{11}{2}\right)$
$\displaystyle =-3+\frac{11}{2}=\frac{5}{2}$
$\displaystyle \text{From formula, nth term,}$
$\displaystyle a_{n}=a+(n-1)d$
$\displaystyle \Rightarrow \frac{49}{2}=-\frac{11}{2}+(n-1)\frac{5}{2}$
$\displaystyle \Rightarrow \frac{60}{2}\times\frac{2}{5}=n-1$
$\displaystyle \Rightarrow n-1=12\Rightarrow n=13$
$\displaystyle \therefore \text{13th term of A.P. is }\frac{49}{2}$
$\\$

$\displaystyle \textbf{Question 16. }\text{Find }a\text{ and }b\text{ so that the numbers }a,7,b,23\text{ are in AP.} \hspace{0.2cm}\text{[CBSE 2022]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }a,\ 7,\ b,\ 23\text{ are in A.P.}$
$\displaystyle \text{So }7-a=b-7$
$\displaystyle \Rightarrow a+b=14 \qquad (i)$
$\displaystyle \text{Also }b-7=23-b$
$\displaystyle \Rightarrow 2b=30\Rightarrow b=15$
$\displaystyle \text{Putting the value of }b\text{ in eq. }(i),\text{ we get}$
$\displaystyle a+15=14$
$\displaystyle \Rightarrow a=-1$
$\\$

$\displaystyle \textbf{Question 17. }\text{Find the sum of first }20\text{ terms of an AP, whose }n\text{th term is given as} \\ a_{n}=5-2n. \hspace{0.2cm}\text{[CBSE 2022]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{nth term of the given A.P. is}$
$\displaystyle a_n=5-2n$
$\displaystyle \therefore \text{1st term}=5-2\times1=3$
$\displaystyle \text{2nd term}=5-2\times2=1$
$\displaystyle \text{3rd term}=5-2\times3=-1$
$\displaystyle \text{So, A.P. is }3,1,-1,\ldots$
$\displaystyle \therefore \text{Common difference }(d)=1-3=-2$
$\displaystyle \text{Now, sum of first }20\text{ terms}$
$\displaystyle =\frac{n}{2}[2a+(n-1)d]$
$\displaystyle =\frac{20}{2}[2\times3+(20-1)\times(-2)]$
$\displaystyle =-320$
$\\$

$\displaystyle \textbf{Question 18. }\text{The first term of an AP is }p\text{ and the common difference is }q,\text{ then its }10^{\text{th}}\text{ term is} \hspace{0.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{(a) }q+9p\qquad \text{(b) }p-9q\qquad \text{(c) }p+9q\qquad \text{(d) }2p+9q$
$\displaystyle \text{Answer:}$
$\displaystyle \text{In an A.P., }a_n=a+(n-1)d$
$\displaystyle \text{Here }a=p,\ d=q,\ n=10$
$\displaystyle \therefore a_{10}=p+(10-1)q$
$\displaystyle =p+9q$
$\displaystyle \therefore \text{Correct option is (c).}$
$\\$

$\displaystyle \textbf{Question 19. }\text{Show that }(a-b)^{2},(a^{2}+b^{2})\text{ and }(a+b)^{2}\text{ are in AP.} \hspace{0.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{If }(a-b)^{2},\ (a^{2}+b^{2})\text{ and }(a+b)^{2}\text{ are in A.P., then}$
$\displaystyle (a^{2}+b^{2})-(a-b)^{2}=(a+b)^{2}-(a^{2}+b^{2})$
$\displaystyle \Rightarrow a^{2}+b^{2}-a^{2}-b^{2}+2ab=a^{2}+b^{2}+2ab-a^{2}-b^{2}$
$\displaystyle \Rightarrow \text{L.H.S.}=\text{R.H.S.}$
$\displaystyle \text{Hence, the given terms are consecutive terms of an A.P.}$
$\\$

$\displaystyle \textbf{Question 20. }\text{The sum of four consecutive numbers in AP is }32\text{ and the ratio of the}$
$\displaystyle \text{product of the first and last terms to the product of two middle terms is }$
$\displaystyle 7:15. \ \text{Find the numbers.} \hspace{0.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the four consecutive numbers of the A.P. are }a-3d,\ a-d,\ a+d\text{ and }a+3d\text{ respectively.}$
$\displaystyle \text{A.T.Q.}$
$\displaystyle a-3d+a-d+a+d+a+3d=32$
$\displaystyle \Rightarrow 4a=32$
$\displaystyle \Rightarrow a=8$
$\displaystyle \text{Also,}$
$\displaystyle \frac{(a-3d)(a+3d)}{(a-d)(a+d)}=\frac{7}{15}$
$\displaystyle \Rightarrow \frac{a^{2}-9d^{2}}{a^{2}-d^{2}}=\frac{7}{15}$
$\displaystyle \Rightarrow \frac{64-9d^{2}}{64-d^{2}}=\frac{7}{15}$
$\displaystyle \Rightarrow 15a^{2}-135d^{2}=7a^{2}-7d^{2}$
$\displaystyle \Rightarrow 8a^{2}=128d^{2}$
$\displaystyle \Rightarrow 8\times8^{2}=128d^{2} \qquad [\because a=8]$
$\displaystyle \Rightarrow 128d^{2}=8\times64$
$\displaystyle \Rightarrow d=\sqrt{4}=\pm2$
$\displaystyle \text{When }d=2\text{ and }a=8,$
$\displaystyle a-3d=8-3\times2=2$
$\displaystyle a-d=8-2=6$
$\displaystyle a+d=8+2=10$
$\displaystyle a+3d=8+3\times2=14$
$\displaystyle \text{When }d=-2\text{ and }a=8,$
$\displaystyle a-3d=8-3(-2)=14$
$\displaystyle a-d=8-(-2)=10$
$\displaystyle a+d=8+(-2)=6$
$\displaystyle a+3d=8+3(-2)=2$
$\displaystyle \therefore \text{Four consecutive terms of the A.P. are }2,6,10,14\text{ or }14,10,6,2\text{ respectively.}$
$\\$

$\displaystyle \textbf{Question 21. }\text{Find the sum of first }20\text{ terms of the following AP : }1,4,7,10,\ldots \\ \hspace{0.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{A.P. is }1,4,7,10,\ldots$
$\displaystyle \therefore a=1,\ d=3$
$\displaystyle \text{Sum of first }20\text{ terms,}$
$\displaystyle S_{20}=\frac{20}{2}[2\times1+(20-1)3]$
$\displaystyle =590$
$\\$

$\displaystyle \textbf{Question 22. }\text{The sum of the first }7\text{ terms of an AP is }63\text{ and that of its next } \\ 7\text{ terms is }161.\text{ Find the AP.} \hspace{0.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be first term, }d\text{ be common difference of an A.P.}$
$\displaystyle S_7=63\text{ and }S_{14}=224$
$\displaystyle \frac{7}{2}[2a+6d]=63$
$\displaystyle \Rightarrow a+3d=9 \qquad (i)$
$\displaystyle \frac{14}{2}[2a+13d]=224$
$\displaystyle \Rightarrow 2a+13d=32 \qquad (ii)$
$\displaystyle \text{Solving }(i)\text{ and }(ii),$
$\displaystyle a=3,\ d=2$
$\displaystyle \therefore \text{A.P. is }3,5,7,9,\ldots$
$\\$

$\displaystyle \textbf{Question 23. }\text{Solve: }1+4+7+10+\cdots+x=287. \hspace{0.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Consider }1+4+7+10+\ldots+x=287$
$\displaystyle \text{Let }a_n=x,\ a=1,\ d=3$
$\displaystyle S_n=287=\frac{n}{2}[2\times1+(n-1)3]$
$\displaystyle \Rightarrow 287=\frac{n}{2}[2+3n-3]$
$\displaystyle \Rightarrow 574=3n^2-n$
$\displaystyle \Rightarrow 3n^2-n-574=0$
$\displaystyle \Rightarrow 3n^2-42n+41n-574=0$
$\displaystyle \Rightarrow 3n(n-14)+41(n-14)=0$
$\displaystyle \Rightarrow (3n+41)(n-14)=0$
$\displaystyle \Rightarrow 3n+41=0\text{ or }n-14=0$
$\displaystyle \Rightarrow n=-\frac{41}{3}\text{ (rejected)}\text{ or }n=14$
$\displaystyle \therefore a_{14}=x=1+(14-1)3$
$\displaystyle \Rightarrow 1+39=x$
$\displaystyle \Rightarrow x=40$
$\\$

$\displaystyle \textbf{Question 24. }\text{How many two digits numbers are divisible by }3? \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Two digit numbers divisible by }3\text{ are }12,15,\ldots,99$
$\displaystyle \text{Here, }a=12,\ d=3,\ a_{n}=99$
$\displaystyle \text{Let two digit numbers divisible by }3\text{ be }n$
$\displaystyle a_{n}=a+(n-1)d$
$\displaystyle \Rightarrow 99=12+(n-1)3$
$\displaystyle \Rightarrow 99=9+3n$
$\displaystyle \Rightarrow 3n=90\Rightarrow n=30$
$\displaystyle \therefore 30\text{ two digit numbers are divisible by }3.$
$\\$

$\displaystyle \textbf{Question 25. }\text{Which term of the AP: }3,15,27,39,\ldots\text{ will be }120 \text{ more than} \\ \text{its }21^{\mathrm{st}}\text{ term?} \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let nth term of the A.P. }3,15,27,39,\ldots\text{ be }120\text{ more than its 21st term.}$
$\displaystyle a_{n}=a_{21}+120$
$\displaystyle a+(n-1)d=a+20d+120$
$\displaystyle \text{(where }a\text{ is 1st term and }d\text{ is common difference)}$
$\displaystyle \Rightarrow (n-1)d=20d+120 \qquad (i)$
$\displaystyle \therefore d=15-3=12$
$\displaystyle \text{Put in eq. }(i),\text{ we get}$
$\displaystyle (n-1)12=20(12)+120$
$\displaystyle \Rightarrow 12n-12=240+120$
$\displaystyle \Rightarrow 12n=360+12$
$\displaystyle \Rightarrow n=\frac{372}{12}=31$
$\displaystyle \therefore \text{31st term is 120 more than its 21st term.}$
$\\$

$\displaystyle \textbf{Question 26. }\text{Which term of the Arithmetic Progression }-7,-12,-17,-22,\ldots$
$\displaystyle \text{ will be }-82.\text{ Is }-100\text{ any term of} \text{ the A.P.? Give reason for your answer.} \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let nth term is }-82,\text{ i.e. }a_n=-82$
$\displaystyle \text{Given: }-7,-12,-17,-22,\ldots$
$\displaystyle \text{Here, }a=-7,\ d=-12+7=-5$
$\displaystyle \therefore a+(n-1)d=-82$
$\displaystyle \Rightarrow -7+(n-1)(-5)=-82$
$\displaystyle \Rightarrow n-1=15 \Rightarrow n=16$
$\displaystyle \therefore 16^{\text{th}}\text{ term is }-82.$
$\displaystyle \text{For checking }(-100),$
$\displaystyle a+(n-1)d=-100$
$\displaystyle \Rightarrow -7+(n-1)(-5)=-100$
$\displaystyle \Rightarrow n-1=\frac{93}{5}=\frac{93}{5}+1$
$\displaystyle \Rightarrow n=\frac{98}{5}=19\frac{3}{5}$
$\displaystyle \text{Which is not an integral number.}$
$\displaystyle \therefore -100\text{ is not a term of the A.P.}$
$\\$

$\displaystyle \textbf{Question 27. }\text{If the sum of first }n\text{ terms of an AP is }n^{2},\text{ then find its }10\text{th term.} \\ \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Here, }S_n\text{ is sum of }n\text{ terms of an A.P.}$
$\displaystyle \therefore S_n=n^2 \qquad (\text{given})$
$\displaystyle \text{nth term of the A.P. is given by}$
$\displaystyle a_n=S_n-S_{n-1}$
$\displaystyle \Rightarrow a_n=n^2-(n-1)^2$
$\displaystyle \Rightarrow a_n=n^2-(n^2+1-2n)$
$\displaystyle \Rightarrow a_n=2n-1$
$\displaystyle \text{Put }n=10$
$\displaystyle \therefore a_{10}=2(10)-1=19$
$\displaystyle \therefore \text{10th term is }19.$
$\\$

$\displaystyle \textbf{Question 28. }\text{If the sum of first four terms of an AP is }40\text{ and that of first }14$
$\displaystyle \text{terms is }280. \text{Find the sum of its first }n\text{ terms.} \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let first term of an A.P. is }a\text{ and common difference is }d.$
$\displaystyle S_4=40$
$\displaystyle \frac{4}{2}[2a+(4-1)d]=40$
$\displaystyle \Rightarrow 2a+3d=20 \qquad (i)$
$\displaystyle \text{and }S_{14}=280$
$\displaystyle \frac{14}{2}[2a+(14-1)d]=280$
$\displaystyle \Rightarrow 2a+13d=40 \qquad (ii)$
$\displaystyle \text{Equation }(ii)-\text{equation }(i)$
$\displaystyle 10d=20\Rightarrow d=2$
$\displaystyle \text{Put }d=2\text{ in equation }(i),\text{ we have}$
$\displaystyle 2a+3(2)=20$
$\displaystyle \Rightarrow 2a=14\Rightarrow a=7$
$\displaystyle \text{Sum of }n\text{ terms}=\frac{n}{2}[14+(n-1)2]$
$\displaystyle =\frac{n}{2}[2n+12]=n(n+6)$
$\\$

$\displaystyle \textbf{Question 29. }\text{The first term of an AP is }3,\text{ the last term is }83\text{ and the sum}$
$\displaystyle \text{of all its terms is }903. \text{Find the number of terms and the common difference}$
$\displaystyle \text{of the AP.} \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Here }a=3\text{ and }a_n=83.$
$\displaystyle \text{Let number of terms be }n.$
$\displaystyle S_n=903$
$\displaystyle S_n=\frac{n}{2}(a+a_n)$
$\displaystyle \Rightarrow 903=\frac{n}{2}(3+83)$
$\displaystyle \Rightarrow n=\frac{1806}{86}=21$
$\displaystyle \therefore a_n=a+(n-1)d$
$\displaystyle 83=3+20d$
$\displaystyle \Rightarrow 80=20d$
$\displaystyle \Rightarrow d=4$
$\displaystyle \text{Here}$
$\displaystyle a=3\text{ and }a_n=83.$
$\displaystyle \text{Let number of terms be }n.$
$\displaystyle S_n=903$
$\displaystyle S_n=\frac{n}{2}(a+a_n)$
$\displaystyle \Rightarrow 903=\frac{n}{2}[3+83]$
$\displaystyle \Rightarrow n=\frac{1806}{86}=21$
$\displaystyle \therefore a_n=a+(n-1)d$
$\displaystyle 83=3+(20)d$
$\displaystyle \Rightarrow 80=20d$
$\displaystyle \Rightarrow d=4$
$\\$

$\displaystyle \textbf{Question 30. }\text{In an AP, if the common difference }(d)=-4\text{ and the seventh} \\ \text{term }(a_{7})\text{ is }4,\text{ then find the first term.} \hspace{0.2cm}\text{[CBSE 2018]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let, }a=\text{ first term,}$
$\displaystyle \text{Given, }d=-4,\ a_{7}=4$
$\displaystyle \Rightarrow a+(7-1)d=4$
$\displaystyle \left[\because \text{nth term of an A.P. }a_{n}=a+(n-1)d\right]$
$\displaystyle \Rightarrow a+6d=4$
$\displaystyle \Rightarrow a+6(-4)=4$
$\displaystyle \Rightarrow a-24=4$
$\displaystyle \Rightarrow a=4+24=28$
$\displaystyle \therefore \text{First term, }a=28.$
$\\$

$\displaystyle \textbf{Question 31. }\text{Find the sum of first }8\text{ multiples of }3. \hspace{0.2cm}\text{[CBSE 2018]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The first 8 multiples of 3 are }3,6,9,12,15,18,21,24.$
$\displaystyle \text{Here, 2nd term - 1st term = 3rd term - 2nd term}$
$\displaystyle \Rightarrow 6-3=9-6=3$
$\displaystyle \text{Since, common difference is same. Therefore, the above terms are in A.P.}$
$\displaystyle \text{Here, }a=3,\ d=6-3=3,\ n=8$
$\displaystyle \text{We know, the sum of }n\text{ terms of an A.P.}$
$\displaystyle S_n=\frac{n}{2}[2a+(n-1)d]$
$\displaystyle \text{where }a=\text{first term, }d=\text{common difference}$
$\displaystyle S_8=\frac{8}{2}[2\times3+(8-1)3]$
$\displaystyle =4[6+7\times3]$
$\displaystyle =4[6+21]$
$\displaystyle =108$
$\\$

$\displaystyle \textbf{Question 32. }\text{In an A.P. if sum of its first }n\text{ terms is }3n^{2}+5n\text{ and its}$
$\displaystyle k^{\mathrm{th}}\text{ term is }164,\text{ find the value of }k. \hspace{0.2cm}\text{[CBSE 2018(C)]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The sum of first }n\text{ terms is }3n^2+5n$
$\displaystyle S_n=3n^2+5n$
$\displaystyle \text{Change }n\text{ to }n-1$
$\displaystyle S_{n-1}=3(n-1)^2+5(n-1)$
$\displaystyle =3n^2-n-2$
$\displaystyle \text{nth term}=S_n-S_{n-1}$
$\displaystyle \text{So, kth term}=6k+2$
$\displaystyle 164=6k+2$
$\displaystyle 162=6k\Rightarrow k=27$
$\\$

$\displaystyle \textbf{Question 33. }\text{What is the common difference of an A.P. in which } \\ a_{21}-a_{7}=84? \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }d\text{ be the common difference of the A.P. whose first term is }a.$
$\displaystyle \text{Now, }a_{21}-a_{7}=84$
$\displaystyle \Rightarrow (a+20d)-(a+6d)=84$
$\displaystyle \Rightarrow d=6$
$\\$

$\displaystyle \textbf{Question 34. }\text{Find how many integers between }200\text{ and }500\text{ are} \\ \text{divisible by }8. \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Numbers divisible by }8\text{ between }200\text{ and }500\text{ are }208,216,224,232,\ldots,496$
$\displaystyle \text{Since, }216-208=224-216=8$
$\displaystyle \therefore \text{The given integers are in A.P.}$
$\displaystyle \text{Let first term }(a)=208,$
$\displaystyle \text{Common difference }(d)=224-216=8,$
$\displaystyle \text{Number of terms }(n)=?$
$\displaystyle \text{Last term }(a_n)=496$
$\displaystyle \text{We know that, }a_n=a+(n-1)d$
$\displaystyle \Rightarrow 496=208+(n-1)(8)$
$\displaystyle \Rightarrow 496-208=(n-1)(8)$
$\displaystyle \Rightarrow n=\frac{296}{8}+1=37$
$\displaystyle \therefore \text{Between }200\text{ and }500\text{ there are }37\text{ integers divisible by }8.$
$\\$

$\displaystyle \textbf{Question 35. }\text{Which term of the progression }20,19\frac{1}{4},18\frac{1}{2},17\frac{3}{4},\ldots \\ \text{is the first negative term?} \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) }20,\ 19\frac{1}{4},\ 18\frac{1}{2},\ 17\frac{3}{4},\ldots$
$\displaystyle \text{First term is }20$
$\displaystyle \text{Also, Second term - First term}=19\frac{1}{4}-20=-\frac{3}{4}$
$\displaystyle \text{Third term - Second term}=18\frac{1}{2}-19\frac{1}{4}=-\frac{3}{4}$
$\displaystyle \text{Fourth term - Third term}=17\frac{3}{4}-18\frac{1}{2}=-\frac{3}{4}$
$\displaystyle \therefore \text{Difference between consecutive terms is constant, therefore this progression is an A.P.}$
$\displaystyle \text{Let nth term be the first negative term.}$
$\displaystyle a_n=a+(n-1)d$
$\displaystyle \Rightarrow a_n=20+(n-1)\left(-\frac{3}{4}\right)$
$\displaystyle \Rightarrow a_n=20-\frac{3}{4}(n-1)$
$\displaystyle \text{Now A.T.Q. }a_n<0$
$\displaystyle \Rightarrow 20-\frac{3}{4}(n-1)<0$
$\displaystyle \Rightarrow 20-\frac{3n}{4}+\frac{3}{4}<0$
$\displaystyle \Rightarrow \frac{83}{4}-\frac{3n}{4}<0$
$\displaystyle \Rightarrow 83-3n<0$
$\displaystyle \Rightarrow n>\frac{83}{3}=27\frac{2}{3}$
$\displaystyle \Rightarrow n\geq 28 \qquad [\because n\text{ is a natural number}]$
$\displaystyle \therefore \text{28th term of the given progression is the first negative term.}$
$\\$

$\displaystyle \textbf{Question 36. }\text{For what value of }n,\text{ are the }n\text{th terms of two A.P.s }63,65,67,\ldots \\ \text{and }3,10,17,\ldots\text{ equal?} \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the first A.P. }63,65,67,\ldots$
$\displaystyle \text{We have }a=63,\ d=2$
$\displaystyle \therefore \text{nth term }a_n=a+(n-1)d$
$\displaystyle \Rightarrow a_n=63+(n-1)2$
$\displaystyle \Rightarrow a_n=61+2n \qquad (i)$
$\displaystyle \text{For the second A.P. }3,10,17,\ldots$
$\displaystyle \text{We have }a'=3,\ d'=7$
$\displaystyle \therefore a'_n=3+(n-1)7$
$\displaystyle \Rightarrow a'_n=7n-4 \qquad (ii)$
$\displaystyle \text{A.T.Q. }61+2n=7n-4$
$\displaystyle \Rightarrow 65=5n$
$\displaystyle \Rightarrow n=13$
$\\$

$\displaystyle \textbf{Question 37. }\text{If seven times the }7^{\mathrm{th}}\text{ term of an A.P. is equal to eleven times the }$
$\displaystyle 11^{\mathrm{th}}\text{ term, then what will be its }18^{\mathrm{th}}\text{ term?} \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let 1st term be }a\text{ and common difference of A.P. is }d$
$\displaystyle \text{Also, }7a_7=11a_{11}$
$\displaystyle \Rightarrow 7(a+6d)=11(a+10d)$
$\displaystyle \Rightarrow a+17d=0$
$\displaystyle \therefore a_{18}=a+17d=0$
$\displaystyle \text{Hence, 18th term of this A.P. is }0.$
$\\$

$\displaystyle \textbf{Question 38. }\text{Write the }n^{\mathrm{th}}\text{ term of the A.P. }\frac{1}{m},\frac{1+m}{m},\frac{1+2m}{m},\ldots \hspace{0.2cm}\text{[CBSE 2017(C)]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{In sequence } \frac{1}{m},\ \frac{1+m}{m},\ \frac{1+2m}{m},\ldots$
$\displaystyle a_1=\frac{1}{m},\ a_2=\frac{1+m}{m},\ a_3=\frac{1+2m}{m}$
$\displaystyle d=a_3-a_2$
$\displaystyle =\frac{1+2m}{m}-\frac{1+m}{m}=\frac{m}{m}=1$
$\displaystyle \text{nth term of the A.P. is}$
$\displaystyle a_n=a_1+(n-1)d$
$\displaystyle =\frac{1}{m}+(n-1)(1)$
$\displaystyle =\frac{1+(n-1)m}{m}$
$\\$

$\displaystyle \textbf{Question 39. }\text{If the }n\text{th term of the A.P. }-1,4,9,14,\ldots\text{ is }129, \text{ find the value} \\ \text{of }n. \hspace{0.2cm}\text{[CBSE 2017(C)]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{A.P. is }-1,4,9,14,\ldots$
$\displaystyle \text{Common difference, }d=4-(-1)=5$
$\displaystyle T_n=129$
$\displaystyle T_n=a+(n-1)d$
$\displaystyle \Rightarrow 129=-1+(n-1)5$
$\displaystyle \Rightarrow 129=-1+5n-5$
$\displaystyle \Rightarrow 135=5n$
$\displaystyle \Rightarrow n=27$
$\\$

$\displaystyle \textbf{Question 40. }\text{If the }p\text{th term of an A.P. is }q\text{ and }q\text{th term is }p,\text{ prove that}$
$\displaystyle \text{its }n\text{th term is }(p+q-n). \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be the first term and }d\text{ be the common difference of the given A.P.}$
$\displaystyle \text{Then}$
$\displaystyle p^{\text{th}}\text{ term}=q\Rightarrow a+(p-1)d=q \qquad (i)$
$\displaystyle q^{\text{th}}\text{ term}=p\Rightarrow a+(q-1)d=p \qquad (ii)$
$\displaystyle \text{Subtracting equation }(i)\text{ from equation }(ii),\text{ we get}$
$\displaystyle (p-1)d-(q-1)d=q-p$
$\displaystyle \Rightarrow (p-q)d=q-p$
$\displaystyle \Rightarrow (p-q)d=-(p-q)$
$\displaystyle \Rightarrow d=-1$
$\displaystyle \text{Putting }d=-1\text{ in equation }(i),\text{ we get}$
$\displaystyle a+(p-1)(-1)=q$
$\displaystyle \Rightarrow a-(p-1)=q$
$\displaystyle \Rightarrow a=q+(p-1)$
$\displaystyle \therefore \text{nth term}=a+(n-1)d$
$\displaystyle =(q+p-1)+(n-1)(-1)$
$\displaystyle =q+p-1-n+1$
$\displaystyle =q+p-n$
$\displaystyle \therefore \text{Hence, the nth term is }p+q-n.$
$\\$

$\displaystyle \textbf{Question 41. }\text{If }m\text{th term of an AP is }\frac{1}{n}\text{ and }n\text{th term is }\frac{1}{m},\text{ then find the sum} \\ \text{of its first }mn\text{ terms.} \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given }a_m=\frac{1}{n}\text{ and }a_n=\frac{1}{m}$
$\displaystyle \Rightarrow a+(m-1)d=\frac{1}{n} \qquad (i)$
$\displaystyle \Rightarrow a+(n-1)d=\frac{1}{m} \qquad (ii)$
$\displaystyle \text{Subtracting }(ii)\text{ from }(i),$
$\displaystyle (m-1)d-(n-1)d=\frac{1}{n}-\frac{1}{m}$
$\displaystyle \Rightarrow d=\frac{1}{mn}$
$\displaystyle \text{From }(i),$
$\displaystyle a+(m-1)\cdot\frac{1}{mn}=\frac{1}{n}$
$\displaystyle \Rightarrow a=\frac{1}{n}-\frac{m-1}{mn}=\frac{1}{mn}$
$\displaystyle S_{mn}=\frac{mn}{2}[2a+(mn-1)d]$
$\displaystyle =\frac{mn}{2}\left[2\cdot\frac{1}{mn}+(mn-1)\cdot\frac{1}{mn}\right]$
$\displaystyle =\frac{mn}{2}\left[\frac{2}{mn}+\frac{mn-1}{mn}\right]$
$\displaystyle =\frac{mn}{2}\left[\frac{mn+1}{mn}\right]$
$\displaystyle =\frac{1}{2}(1+mn)$
$\\$

$\displaystyle \textbf{Question 42. }\text{Find the sum of }n\text{ terms of the series}$
$\displaystyle \left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+\cdots \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given series is }(4-\frac{1}{n})+(4-\frac{2}{n})+(4-\frac{3}{n})+\ldots$
$\displaystyle S_n=(4+4+4+\ldots n\text{ times})-\left(\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+\ldots+\frac{n}{n}\right)$
$\displaystyle S_n=4n-\left(\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+\ldots+\frac{n}{n}\right) \qquad (i)$
$\displaystyle \text{To find sum of }\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+\ldots+\frac{n}{n}$
$\displaystyle \text{Taking }\frac{1}{n}\text{ common, the series becomes}$
$\displaystyle \frac{1}{n}(1+2+3+\ldots+n)$
$\displaystyle \text{Here }a=1,\ l=n$
$\displaystyle \Rightarrow \frac{1}{n}\times\frac{n}{2}(1+n)\left[\text{Using }S_n=\frac{n}{2}(a+l)\right]$
$\displaystyle =\frac{1+n}{2}$
$\displaystyle \therefore (i)\text{ becomes,}$
$\displaystyle S_n=4n-\left(\frac{1+n}{2}\right)$
$\displaystyle =\frac{8n-1-n}{2}=\frac{1}{2}(7n-1)$
$\\$

$\displaystyle \textbf{Question 43. }\text{The ratio of the sums of first }m\text{ and first }n\text{ terms of an AP is }$
$\displaystyle m^{2}:n^{2}.\text{ Show that the ratio} \text{of its }m\text{th and }n\text{th terms is }(2m-1):(2n-1). \\ \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let 1st term of the A.P. be }a\text{ and common difference be }d.$
$\displaystyle S_m=\frac{m}{2}[2a+(m-1)d]$
$\displaystyle S_n=\frac{n}{2}[2a+(n-1)d]$
$\displaystyle \text{A.T.Q.}$
$\displaystyle \frac{S_m}{S_n}=\frac{m^2}{n^2}$
$\displaystyle \Rightarrow \frac{\frac{m}{2}[2a+(m-1)d]}{\frac{n}{2}[2a+(n-1)d]}=\frac{m^2}{n^2}$
$\displaystyle \Rightarrow \frac{2a+(m-1)d}{2a+(n-1)d}=\frac{m^2}{n^2}\times\frac{n}{m}=\frac{m}{n}$
$\displaystyle \text{Replacing }m\text{ by }2m-1\text{ and }n\text{ by }2n-1,\text{ we get}$
$\displaystyle \frac{2a+(2m-1-1)d}{2a+(2n-1-1)d}=\frac{2m-1}{2n-1}$
$\displaystyle \Rightarrow \frac{a+(m-1)d}{a+(n-1)d}=\frac{2m-1}{2n-1}$
$\displaystyle \Rightarrow \frac{a_m}{a_n}=\frac{2m-1}{2n-1}$
$\displaystyle \therefore \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 44. }\text{How many terms of AP }3,9,17,25,\ldots\text{ must be taken to give} \\ \text{a sum of }636? \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the given A.P. }9,17,25,\ldots$
$\displaystyle a=9,\ d=8$
$\displaystyle \text{Given sum }S_n=636$
$\displaystyle \text{We know that}$
$\displaystyle S_n=\frac{n}{2}[2a+(n-1)d]$
$\displaystyle \therefore 636=\frac{n}{2}[2\times9+(n-1)8]$
$\displaystyle \Rightarrow 636=\frac{n}{2}[18+8n-8]$
$\displaystyle \Rightarrow 636=\frac{n}{2}[10+8n]$
$\displaystyle \Rightarrow 636\times2=n[10+8n]$
$\displaystyle \Rightarrow 636=5n+4n^{2}$
$\displaystyle \Rightarrow 4n^{2}+5n-636=0$
$\displaystyle \Rightarrow 4n^{2}-48n+53n-636=0$
$\displaystyle \Rightarrow 4n(n-12)+53(n-12)=0$
$\displaystyle \Rightarrow (n-12)(4n+53)=0$
$\displaystyle \Rightarrow n=-\frac{53}{4},\ n=12$
$\displaystyle n=-\frac{53}{4}\text{ is rejected because }n\text{ has to be a natural number.}$
$\displaystyle \therefore n=12.$
$\\$

$\displaystyle \textbf{Question 45. }\text{The first term of an A.P. is }5,\text{ the last term is }45\text{ and the sum of all its terms is }$
$\displaystyle 400.\text{ Find the} \text{number of terms and the common difference of the A.P.} \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be the first term and }d\text{ be the common difference of the given A.P.}$
$\displaystyle \text{Let }a_n\text{ be the last term and }S_n\text{ be the sum of }n\text{ terms.}$
$\displaystyle \text{So, }a=5,\ a_n=45,\ S_n=400$
$\displaystyle \text{We need to find the values of }n\text{ and }d.$
$\displaystyle S_n=\frac{n}{2}(a+a_n) \qquad (i)$
$\displaystyle \text{Substituting the values in }(i)$
$\displaystyle 400=\frac{n}{2}(5+45)$
$\displaystyle \Rightarrow \frac{400\times2}{50}=n$
$\displaystyle \Rightarrow n=16$
$\displaystyle \text{Also,}$
$\displaystyle a_n=a+(n-1)d$
$\displaystyle 45=5+(16-1)d$
$\displaystyle \Rightarrow 45-5=15d$
$\displaystyle \Rightarrow d=\frac{40}{15}=\frac{8}{3}$
$\displaystyle \therefore \text{There are }16\text{ terms and common difference is }\frac{8}{3}.$
$\\$

$\displaystyle \textbf{Question 46. }\text{Find the sum of the following series:}$
$\displaystyle 5+(-41)+9+(-39)+13+(-37)+17+\cdots+(-5)+81+(-3) \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The series can be rewritten as,}$
$\displaystyle (5+9+13+\ldots+81)+\{-41+(-39)+(-37)+\ldots+(-5)+(-3)\}$
$\displaystyle \text{For }5+9+13+\ldots+81$
$\displaystyle a=5,\ d=4$
$\displaystyle a_n=81$
$\displaystyle 5+(n-1)4=81 \Rightarrow n=20$
$\displaystyle S_n=\frac{20}{2}(5+81)=860$
$\displaystyle \text{For }(-41)+(-39)+(-37)+\ldots+(-5)+(-3)$
$\displaystyle a=-41,\ d=2,\ a_n=-3$
$\displaystyle -41+(n-1)2=-3$
$\displaystyle \Rightarrow n=20$
$\displaystyle S_n=\frac{20}{2}(-41-3)=-440$
$\displaystyle \text{Sum of series}=860-440=420$
$\\$

$\displaystyle \textbf{Question 47. }\text{In an A.P. of }50\text{ terms, the sum of the first }10\text{ terms is }210$
$\displaystyle \text{ and the sum of its last }15\text{ terms is }2565.\text{ Find } \text{the A.P.} \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given }S_{10}=210$
$\displaystyle \Rightarrow \frac{10}{2}(2a+9d)=210$
$\displaystyle \Rightarrow 2a+9d=42 \qquad (i)$
$\displaystyle a_{36}=a+35d,\quad a_{50}=a+49d$
$\displaystyle \text{Sum of last 15 terms}=\frac{15}{2}(a+35d+a+49d)$
$\displaystyle 2565=\frac{15}{2}(2a+84d)$
$\displaystyle \Rightarrow a+42d=171 \qquad (ii)$
$\displaystyle \text{Solving }(i)\text{ and }(ii),\text{ we get}$
$\displaystyle a=3,\ d=4$
$\displaystyle a_1=3,\ a_2=7,\ a_3=11,\ a_4=15,\ldots$
$\displaystyle \therefore \text{A.P. is }3,7,11,\ldots$
$\\$

$\displaystyle \textbf{Question 48. }\text{If the ratio of the sum of the first }n\text{ terms of two A.P.s is }$
$\displaystyle (7n+1):(4n+27),\text{ then find the ratio }\text{of their }9^{\mathrm{th}}\text{ terms.} \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a_1,a_2\text{ be the first terms and }d_1,d_2\text{ be the common differences of the two given A.P.'s.}$
$\displaystyle \text{Then the sums of their first }n\text{ terms is given by}$
$\displaystyle S_n=\frac{n}{2}[2a_1+(n-1)d_1]$
$\displaystyle \text{and}$
$\displaystyle S'_n=\frac{n}{2}[2a_2+(n-1)d_2]$
$\displaystyle \text{A.T.Q.}$
$\displaystyle \frac{S_n}{S'_n}=\frac{\frac{n}{2}[2a_1+(n-1)d_1]}{\frac{n}{2}[2a_2+(n-1)d_2]}$
$\displaystyle \Rightarrow \frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{7n+1}{4n+27}$
$\displaystyle \Rightarrow \frac{a_1+\frac{(n-1)}{2}d_1}{a_2+\frac{(n-1)}{2}d_2}=\frac{7n+1}{4n+27}$
$\displaystyle \text{Where }\frac{n-1}{2}=8$
$\displaystyle \Rightarrow n=17$
$\displaystyle \therefore \frac{a_1+8d_1}{a_2+8d_2}=\frac{7\times17+1}{4\times17+27}$
$\displaystyle \Rightarrow \frac{a_1+8d_1}{a_2+8d_2}=\frac{120}{95}=\frac{24}{19}$
$\displaystyle \therefore \text{Ratio of 9th terms of these A.P.'s is }24:19.$
$\\$

$\displaystyle \textbf{Question 49. }\text{If the ratio of the }11^{\mathrm{th}}\text{ term of an A.P. to its }18^{\mathrm{th}}\text{ term is }$
$\displaystyle 2:3, \ \text{find the ratio of the sum of the first five terms to the sum of its first }10\text{ terms.}$
$\displaystyle \hspace{0.2cm}\text{[CBSE 2017(C)]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let first term of an A.P. be }a\text{ and common difference }d.$
$\displaystyle \frac{T_{11}}{T_{18}}=\frac{2}{3}$
$\displaystyle \Rightarrow \frac{a+10d}{a+17d}=\frac{2}{3}$
$\displaystyle \Rightarrow 3a+30d=2a+34d$
$\displaystyle \Rightarrow a=4d \qquad (i)$
$\displaystyle \text{So,}$
$\displaystyle \frac{S_5}{S_{10}}=\frac{\frac{5}{2}(2a+4d)}{\frac{10}{2}(2a+9d)}$
$\displaystyle \Rightarrow \frac{S_5}{S_{10}}=\frac{1}{2}\left[\frac{2a+4d}{2a+9d}\right]$
$\displaystyle \text{Put }a=4d$
$\displaystyle \Rightarrow \frac{S_5}{S_{10}}=\frac{1}{2}\left[\frac{2\times4d+4d}{2\times4d+9d}\right]$
$\displaystyle \Rightarrow \frac{S_5}{S_{10}}=\frac{1}{2}\left[\frac{12d}{17d}\right]$
$\displaystyle \Rightarrow \frac{S_5}{S_{10}}=\frac{6}{17}$
$\displaystyle \therefore \text{Required ratio is }6:17.$
$\\$

$\displaystyle \textbf{Question 50. }\text{Find the }9^{\mathrm{th}}\text{ term from the end (towards } \text{the first term) } \\ \text{of the A.P. }5,9,13,\ldots,185. \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Reversing the given A.P. we get}$
$\displaystyle 185,181,177,173,\ldots$
$\displaystyle \text{Now, first term }(a)=185$
$\displaystyle \text{Common difference, }(d)=181-185=-4$
$\displaystyle \text{We know that nth term of an A.P. is given by }a+(n-1)d$
$\displaystyle \text{Ninth term }a_9=a+(9-1)d$
$\displaystyle \Rightarrow a_9=185+8(-4)$
$\displaystyle \Rightarrow a_9=185-32=153$
$\\$

$\displaystyle \textbf{Question 51. }\text{For what value of }k\text{ will }k+9,2k-1\text{ and }2k+7\text{ are the} \\ \text{consecutive terms of an A.P.?} \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }k+9,\ 2k-1\text{ and }2k+7\text{ are in A.P.}$
$\displaystyle \Rightarrow (2k-1)-(k+9)=(2k+7)-(2k-1)$
$\displaystyle \Rightarrow k-10=8$
$\displaystyle \Rightarrow k=18$
$\\$

$\displaystyle \textbf{Question 52. }\text{The }4^{\mathrm{th}}\text{ term of an A.P. is zero. Prove that the }25^{\mathrm{th}}$
$\displaystyle \text{ term of the A.P. is three times its }11^{\mathrm{th}}\text{ term.} \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be first term and }d\text{ be the common difference of the A.P. Then}$
$\displaystyle a_n=a+(n-1)d$
$\displaystyle a_4=a+(4-1)d$
$\displaystyle 0=a+3d \Rightarrow a=-3d \qquad [\because a_4=0]$
$\displaystyle \text{Now}$
$\displaystyle a_{25}=a+(25-1)d$
$\displaystyle =a+24d$
$\displaystyle =-3d+24d=21d=3\times7d$
$\displaystyle \text{Hence,}$
$\displaystyle a_{25}=3\times a_{11} \text{ as } a_{11}=a+10d=-3d+10d=7d.$
$\\$

$\displaystyle \textbf{Question 53. }\text{The digits of a positive number of three digits are in A.P. and their}$
$\displaystyle \text{sum is }15. \text{ The number obtained by reversing the digits is }594\text{ less than the original}$
$\displaystyle \text{number. Find the number.} \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required numbers in A.P. are }a-d,\ a,\ a+d\text{ respectively.}$
$\displaystyle \text{Now, }a-d+a+a+d=15$
$\displaystyle \Rightarrow 3a=15\Rightarrow a=5$
$\displaystyle \text{According to question, number is}$
$\displaystyle 100(a-d)+10a+a+d\text{ i.e. }111a-99d$
$\displaystyle \text{Number on reversing the digits is}$
$\displaystyle 100(a+d)+10a+a-d\text{ i.e. }111a+99d$
$\displaystyle \text{Now, as per given condition in question,}$
$\displaystyle (111a-99d)-(111a+99d)=594$
$\displaystyle \Rightarrow -198d=594$
$\displaystyle \Rightarrow d=-3$
$\displaystyle \therefore \text{Digits of number are }[5-(-3),\ 5,\ 5+(-3)]=8,5,2$
$\displaystyle \therefore \text{Required number is }111\times5-99(-3)=555+297=852$
$\\$

$\displaystyle \textbf{Question 54. }\text{Divide }56\text{ in four parts in A.P. such that the ratio of the product of}$
$\displaystyle \text{their extremes }(1\text{st and }4\text{th})\text{ to the product of means }(2\text{nd and }3\text{rd}) \text{is }5:6. \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the four parts of the A.P. are }a-3d,\ a-d,\ a+d,\ a+3d.$
$\displaystyle \text{Now, }a-3d+a-d+a+d+a+3d=56$
$\displaystyle \Rightarrow 4a=56 \qquad [\because \text{Sum}=56]$
$\displaystyle \Rightarrow a=14$
$\displaystyle \text{According to question,}$
$\displaystyle \frac{(a-3d)(a+3d)}{(a-d)(a+d)}=\frac{5}{6}$
$\displaystyle \Rightarrow \frac{(14-3d)(14+3d)}{(14-d)(14+d)}=\frac{5}{6} \qquad [\because \text{Putting }a=14]$
$\displaystyle \Rightarrow \frac{196-9d^{2}}{196-d^{2}}=\frac{5}{6}$
$\displaystyle \Rightarrow 1176-54d^{2}=980-5d^{2}$
$\displaystyle \Rightarrow d=\pm2$
$\displaystyle \therefore \text{4 parts are }a-3d,\ a-d,\ a+d,\ a+3d\text{ i.e. }8,12,16,20\text{ or }20,16,12,8\text{ respectively.}$
$\\$

$\displaystyle \textbf{Question 55. }\text{The }p\text{th, }q\text{th and }r\text{th terms of an A.P. are }a,b\text{ and }c$
$\displaystyle \text{respectively. Show that }a(q-r)+b(r-p)+c(p-q)=0. \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A\text{ and }d\text{ be the first term and common difference of the given A.P. Then}$
$\displaystyle a_p=A+(p-1)d=a \qquad (i)$
$\displaystyle a_q=A+(q-1)d=b \qquad (ii)$
$\displaystyle a_r=A+(r-1)d=c \qquad (iii)$
$\displaystyle \text{Now, subtracting }(i)\text{ from }(ii),\text{ we get}$
$\displaystyle (p-q)d=a-b$
$\displaystyle \Rightarrow p-q=\frac{a-b}{d}$
$\displaystyle \text{Multiplying by }c\text{ both sides,}$
$\displaystyle c(p-q)=\frac{ca}{d}-\frac{cb}{d} \qquad (iv)$
$\displaystyle \text{Now, }(ii)-(iii),\text{ we get}$
$\displaystyle (q-r)d=b-c$
$\displaystyle \Rightarrow q-r=\frac{b-c}{d}$
$\displaystyle \text{Multiplying by }a\text{ both sides,}$
$\displaystyle a(q-r)=\frac{ab}{d}-\frac{ac}{d} \qquad (v)$
$\displaystyle \text{Now, }(iii)-(i),\text{ we get}$
$\displaystyle (r-p)d=c-a$
$\displaystyle \Rightarrow r-p=\frac{c-a}{d}$
$\displaystyle \text{Multiplying by }b\text{ both sides,}$
$\displaystyle b(r-p)=\frac{bc}{d}-\frac{ab}{d} \qquad (vi)$
$\displaystyle \text{Adding }(iv),\ (v)\text{ and }(vi),\text{ we get}$
$\displaystyle c(p-q)+a(q-r)+b(r-p)=0$
$\\$

$\displaystyle \textbf{Question 56. }\text{How many terms of the AP }18,16,14,\ldots\text{ be taken so} \\ \text{that their sum is zero?} \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the number of terms taken for sum to be zero be }n.$
$\displaystyle \text{Then, sum of }n\text{ terms }(S_n)=0 \qquad (\text{Given})$
$\displaystyle \text{First term }(a)=18$
$\displaystyle \text{Common difference }(d)=-2$
$\displaystyle \text{Therefore,}$
$\displaystyle S_n=\frac{n}{2}[2a+(n-1)d]$
$\displaystyle \Rightarrow 0=\frac{n}{2}[2\times18+(n-1)(-2)]$
$\displaystyle \Rightarrow 0=38-2n$
$\displaystyle \Rightarrow n=19$
$\displaystyle \therefore \text{Sum of 19 terms is }0.$
$\\$

$\displaystyle \textbf{Question 57. }\text{The sums of first }n\text{ terms of three arithmetic progressions are }S_{1},S_{2}$
$\displaystyle \text{and }S_{3}\text{ respectively. The first term of each A.P. is }1\text{ and their common differences}$
$\displaystyle \text{are }1,2\text{ and }3\text{ respectively. Prove that }S_{1}+S_{3}=2S_{2}. \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Here, sum of }n\text{ terms of A.P. is}$
$\displaystyle S_n=\frac{n}{2}[2a+(n-1)d]$
$\displaystyle \therefore S_1=\frac{n}{2}[2+(n-1)1]$
$\displaystyle =\frac{n(n+1)}{2} \qquad [\because a=1,\ d=1]$
$\displaystyle \therefore S_2=\frac{n}{2}[2+(n-1)2]$
$\displaystyle =\frac{n}{2}(2n)=n^{2} \qquad [\because a=1,\ d=2]$
$\displaystyle \therefore S_3=\frac{n}{2}[2+(n-1)3]$
$\displaystyle =\frac{n}{2}[2+3n-3]$
$\displaystyle =\frac{n}{2}(3n-1)$
$\displaystyle \text{Now, consider }S_1+S_3$
$\displaystyle S_1+S_3=\frac{n^{2}+n+3n^{2}-n}{2}$
$\displaystyle =2n^{2}=2S_2$
$\\$

$\displaystyle \textbf{Question 58. }\text{A thief runs with a uniform speed of }100\text{ m/minute. After one minute}$
$\displaystyle \text{a policeman runs after the thief to catch him. He goes with a speed of }100\text{ m/minute}$
$\displaystyle \text{in the first minute and increases his speed by }10\text{ m/minute every succeeding minute.}$
$\displaystyle \text{After how many minutes will the policeman catch the thief?} \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let total time be }(n-1)\text{ minutes in which the police catch the thief.}$
$\displaystyle \text{Since thief ran 1 minute before police start running.}$
$\displaystyle \therefore \text{Time taken by thief before he was caught}=(n-1+1)=n\text{ minute}$
$\displaystyle \text{Then total distance covered by thief}=(100\times n)\text{ metres}$
$\displaystyle \text{Total distance covered by policeman in }(n-1)\text{ minute}$
$\displaystyle =100+110+120+\cdots+(n-1)\text{ terms}$
$\displaystyle =\frac{(n-1)}{2}[200+(n-2)10]$
$\displaystyle \text{According to question,}$
$\displaystyle \text{Total distance covered by thief in }n\text{ minute}=\text{Total distance covered by policeman in }(n-1)\text{ minute}$
$\displaystyle 100n=\frac{(n-1)}{2}[200+10n-20]$
$\displaystyle \Rightarrow 200n=10(n-1)(n+18)$
$\displaystyle \Rightarrow 20n=(n-1)(n+18)$
$\displaystyle \Rightarrow n^2-3n-18=0$
$\displaystyle \Rightarrow n^2-6n+3n-18=0$
$\displaystyle \Rightarrow n(n-6)+3(n-6)=0$
$\displaystyle \Rightarrow (n-6)(n+3)=0$
$\displaystyle \Rightarrow n=6\text{ or }n=-3\text{ (rejected)}$
$\displaystyle \therefore \text{Time taken by policeman to catch the thief is }(6-1)\text{ i.e. }5\text{ minutes.}$
$\\$

$\displaystyle \textbf{Question 59. }\text{The houses in a row are numbered consecutively from }1\text{ to }49.$
$\displaystyle \text{Show that there exists a value of }X\text{ such that sum of numbers of houses preceding}$
$\displaystyle \text{the house numbered }X\text{ is equal to sum of the numbers of houses following }X.$
$\displaystyle \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The A.P. of numbers of houses preceding house numbered }x\text{ is}$
$\displaystyle 1+2+3+\cdots+(x-1)$
$\displaystyle \therefore \text{Sum,}$
$\displaystyle S_n=\frac{n}{2}[2a+(n-1)d],\ \text{where }a\rightarrow \text{first term},\ d\rightarrow \text{common difference}$
$\displaystyle =\frac{(x-1)}{2}[2\times1+(x-1-1)\times1]$
$\displaystyle =\frac{(x-1)}{2}[2+x-2]$
$\displaystyle =\frac{x(x-1)}{2}$
$\displaystyle \text{Now, A.P. of total number of houses following }x\text{ is:}$
$\displaystyle (x+1)+(x+2)+\cdots+49$
$\displaystyle n=49-(x+1)+1=49-x$
$\displaystyle \therefore \text{Sum of these numbers,}$
$\displaystyle S_n=\frac{n}{2}[a+l],\ \text{where }l\text{ is last term}$
$\displaystyle =\frac{(49-x)}{2}[x+1+49]$
$\displaystyle =\frac{(49-x)(x+50)}{2}$
$\displaystyle \text{According to question,}$
$\displaystyle \frac{x(x-1)}{2}=\frac{(49-x)(x+50)}{2}$
$\displaystyle \Rightarrow x^2-x=49x+2450-x^2-50x$
$\displaystyle \Rightarrow 2x^2=2450$
$\displaystyle \Rightarrow x^2=1225\Rightarrow x=35$
$\displaystyle \textbf{Justification:}$
$\displaystyle \text{Now, A.P. of numbers before house numbered }x$
$\displaystyle =1+2+\cdots+34$
$\displaystyle S_{34}=\frac{34}{2}[1+34]$
$\displaystyle =17\times35=595$
$\displaystyle \text{Now, A.P. of numbers following house numbered }x$
$\displaystyle =36+37+\cdots+49$
$\displaystyle S=\frac{14}{2}[36+49]=7\times85=595$
$\displaystyle \text{Hence, for value of }x=35,\text{ the sum of numbers of houses preceding house numbered }x$
$\displaystyle \text{is equal to sum of numbers of houses following }x.$
$\\$

$\displaystyle \textbf{Question 60. }\text{Reshma wanted to save at least Rs }6500\text{ for sending her daughter}$
$\displaystyle \text{to school next year (after }12\text{ months). She saved Rs }450\text{ in the first month and}$
$\displaystyle \text{raised her savings by Rs }20\text{ every next month. How much will she be able to save}$
$\displaystyle \text{in next }12\text{ months? Will she be able to send her daughter to the school next year?}$
$\displaystyle \text{What value is reflected in this question?} \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The amounts saved form an A.P. }450,470,490,\ldots$
$\displaystyle \text{in which}$
$\displaystyle \text{first term }(a)=\text{Rs. }450$
$\displaystyle \text{Common difference }(d)=\text{Rs. }20$
$\displaystyle \text{Total terms }(n)=12\text{ (number of months)}$
$\displaystyle \text{Then,}$
$\displaystyle S_n=\frac{n}{2}[2a+(n-1)d]$
$\displaystyle S_{12}=\frac{12}{2}[2\times450+(12-1)(20)]$
$\displaystyle =6[900+220]=6\times1120=6720$
$\displaystyle \text{Since, }6720>6500$
$\displaystyle \therefore \text{Reshma will be able to send her daughter to school}$
$\displaystyle \text{as she has saved more than Rs. }6500.$
$\displaystyle \text{Now, Reshma is very much concerned about her}$
$\displaystyle \text{daughter's education. She is aware and dedicated}$
$\displaystyle \text{towards her daughter's education.}$
$\\$