Class 10: Arithmetic Progressions

$\displaystyle \textbf{Question 1: }\text{Find:}$
$\displaystyle \text{(i) }10\text{th term of the A.P. }1,4,7,10,\ldots$
$\displaystyle \text{(ii) }18\text{th term of the A.P. }\sqrt2,3\sqrt2,5\sqrt2,\ldots$
$\displaystyle \text{(iii) }n\text{th term of the A.P. }13,8,3,-2,\ldots$
$\displaystyle \text{(iv) }10\text{th term of the A.P. }-40,-15,10,35,\ldots$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know, }a_n=a+(n-1)d.$
$\displaystyle \text{(i) }a=1,\quad d=4-1=3$
$\displaystyle a_{10}=1+(10-1)(3)=1+27=28$
$\displaystyle \therefore \text{The }10\text{th term is }28.$

$\displaystyle \text{(ii) }a=\sqrt2,\quad d=3\sqrt2-\sqrt2=2\sqrt2$
$\displaystyle a_{18}=\sqrt2+(18-1)(2\sqrt2)=\sqrt2+34\sqrt2=35\sqrt2$
$\displaystyle \therefore \text{The }18\text{th term is }35\sqrt2.$

$\displaystyle \text{(iii) }a=13,\quad d=8-13=-5$
$\displaystyle a_n=13+(n-1)(-5)=13-5n+5=18-5n$
$\displaystyle \therefore \text{The }n\text{th term is }18-5n.$

$\displaystyle \text{(iv) }a=-40,\quad d=-15-(-40)=25$
$\displaystyle a_{10}=-40+(10-1)(25)=-40+225=185$
$\displaystyle \therefore \text{The }10\text{th term is }185.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{(i) Which term of the A.P. }3,8,13,\ldots\text{ is }248?$
$\displaystyle \text{(ii) Which term of the A.P. }84,80,76,\ldots\text{ is }0?$
$\displaystyle \text{(iii) Which term of the A.P. }65,61,57,53,\ldots\text{ is its first negative term?}$
$\displaystyle \text{(iv) Which term of the A.P. }-7,-12,-17,-22,\ldots\text{ will be }-82?\text{ Is }-100\text{ any term}$
$\displaystyle \text{of the A.P.?}\hfill\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know, }a_n=a+(n-1)d.$
$\displaystyle \text{(i) }a=3,\quad d=8-3=5$
$\displaystyle 248=3+(n-1)5$
$\displaystyle 245=5(n-1)\Rightarrow n-1=49\Rightarrow n=50$
$\displaystyle \therefore 248\text{ is the }50\text{th term of the A.P.}$

$\displaystyle \text{(ii) }a=84,\quad d=80-84=-4$
$\displaystyle 0=84+(n-1)(-4)$
$\displaystyle 4(n-1)=84\Rightarrow n-1=21\Rightarrow n=22$
$\displaystyle \therefore 0\text{ is the }22\text{nd term of the A.P.}$

$\displaystyle \text{(iii) }a=65,\quad d=61-65=-4$
$\displaystyle a_n=65+(n-1)(-4)=69-4n$
$\displaystyle \text{For the first negative term, }69-4n<0$
$\displaystyle 4n>69\Rightarrow n>\frac{69}{4}=17.25$
$\displaystyle \text{The least natural number satisfying this is }n=18.$
$\displaystyle a_{17}=69-4(17)=1,\quad a_{18}=69-4(18)=-3$
$\displaystyle \therefore \text{The }18\text{th term is the first negative term.}$

$\displaystyle \text{(iv) }a=-7,\quad d=-12-(-7)=-5$
$\displaystyle -82=-7+(n-1)(-5)$
$\displaystyle -75=-5(n-1)\Rightarrow n-1=15\Rightarrow n=16$
$\displaystyle \therefore -82\text{ is the }16\text{th term of the A.P.}$
$\displaystyle \text{For }-100,\quad -100=-7+(n-1)(-5)$
$\displaystyle -93=-5(n-1)\Rightarrow n-1=\frac{93}{5}\Rightarrow n=\frac{98}{5}$
$\displaystyle \text{Since }n\text{ is not a natural number, }-100\text{ is not a term of the A.P.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{(i) Is }302\text{ a term of the A.P. }3,8,13,\ldots?$
$\displaystyle \text{(ii) Is }-150\text{ a term of the A.P. }11,8,5,2,\ldots?\hfill\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know, }a_n=a+(n-1)d.$
$\displaystyle \text{(i) }a=3,\quad d=8-3=5$
$\displaystyle 302=3+(n-1)5$
$\displaystyle 299=5(n-1)\Rightarrow n-1=\frac{299}{5}\Rightarrow n=\frac{304}{5}$
$\displaystyle \text{Since }n\text{ is not a natural number, }302\text{ is not a term of the A.P.}$

$\displaystyle \text{(ii) }a=11,\quad d=8-11=-3$
$\displaystyle -150=11+(n-1)(-3)$
$\displaystyle -161=-3(n-1)\Rightarrow n-1=\frac{161}{3}\Rightarrow n=\frac{164}{3}$
$\displaystyle \text{Since }n\text{ is not a natural number, }-150\text{ is not a term of the A.P.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{How many terms are there in the A.P.?}$
$\displaystyle \text{(i) }7,10,13,\ldots,43\qquad\text{(ii) }-1,-\frac56,-\frac23,-\frac12,\ldots,\frac{10}{3}$
$\displaystyle \text{(iii) }7,13,19,\ldots,205\qquad\text{(iv) }18,15\frac12,13,\ldots,-47$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know, }a_n=a+(n-1)d.$
$\displaystyle \text{(i) }a=7,\quad d=10-7=3,\quad a_n=43$
$\displaystyle 43=7+(n-1)3$
$\displaystyle 36=3(n-1)\Rightarrow n-1=12\Rightarrow n=13$
$\displaystyle \therefore \text{There are }13\text{ terms in the A.P.}$

$\displaystyle \text{(ii) }a=-1,\quad d=-\frac56-(-1)=\frac16,\quad a_n=\frac{10}{3}$
$\displaystyle \frac{10}{3}=-1+(n-1)\frac16$
$\displaystyle \frac{13}{3}=\frac{n-1}{6}\Rightarrow n-1=26\Rightarrow n=27$
$\displaystyle \therefore \text{There are }27\text{ terms in the A.P.}$

$\displaystyle \text{(iii) }a=7,\quad d=13-7=6,\quad a_n=205$
$\displaystyle 205=7+(n-1)6$
$\displaystyle 198=6(n-1)\Rightarrow n-1=33\Rightarrow n=34$
$\displaystyle \therefore \text{There are }34\text{ terms in the A.P.}$

$\displaystyle \text{(iv) }a=18,\quad d=15\frac12-18=-\frac52,\quad a_n=-47$
$\displaystyle -47=18+(n-1)\left(-\frac52\right)$
$\displaystyle -65=-\frac52(n-1)\Rightarrow n-1=26\Rightarrow n=27$
$\displaystyle \therefore \text{There are }27\text{ terms in the A.P.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{The first term of an A.P. is }5,\text{ the common difference is }$
$\displaystyle 3\text{ and the last term is }80; \ \text{find the number of terms.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }a=5,\quad d=3,\quad a_n=80$
$\displaystyle \text{We know, }a_n=a+(n-1)d.$
$\displaystyle 80=5+(n-1)3$
$\displaystyle 75=3(n-1)\Rightarrow n-1=25\Rightarrow n=26$
$\displaystyle \therefore \text{The A.P. has }26\text{ terms.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{The }6\text{th and }17\text{th terms of an A.P. are }19\text{ and }41\text{ respectively; find the }40\text{th term.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know, }a_n=a+(n-1)d.$
$\displaystyle a_6=a+5d=19\qquad\ldots\text{(i)}$
$\displaystyle a_{17}=a+16d=41\qquad\ldots\text{(ii)}$
$\displaystyle \text{Subtracting (i) from (ii), }11d=22\Rightarrow d=2$
$\displaystyle \text{Putting }d=2\text{ in (i), }a+10=19\Rightarrow a=9$
$\displaystyle a_{40}=a+39d=9+39(2)=87$
$\displaystyle \therefore \text{The }40\text{th term is }87.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Find the }12\text{th term from the end of the following arithmetic progressions:}$
$\displaystyle \text{(i) }3,8,13,\ldots,253\ \qquad\text{(ii) }1,4,7,10,\ldots,88$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know, }n\text{th term from the end}=l-(n-1)d.$
$\displaystyle \text{(i) }l=253,\quad d=8-3=5$
$\displaystyle \text{12th term from the end}=253-(12-1)(5)=253-55=198$
$\displaystyle \therefore \text{The }12\text{th term from the end is }198.$

$\displaystyle \text{(ii) }l=88,\quad d=4-1=3$
$\displaystyle \text{12th term from the end}=88-(12-1)(3)=88-33=55$
$\displaystyle \therefore \text{The }12\text{th term from the end is }55.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{If the }5\text{th term of an A.P. is }31\text{ and }25\text{th term is }140\text{ more than the }5\text{th}$
$\displaystyle \text{term, find the A.P.}\hfill\text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know, }a_n=a+(n-1)d.$
$\displaystyle a_5=a+4d=31\qquad\ldots\text{(i)}$
$\displaystyle a_{25}=a_5+140$
$\displaystyle a+24d=a+4d+140$
$\displaystyle 20d=140\Rightarrow d=7$
$\displaystyle \text{Putting }d=7\text{ in (i), }a+28=31\Rightarrow a=3$
$\displaystyle \therefore \text{The A.P. is }3,10,17,24,\ldots$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Find whether }0\text{ (zero) is a term of the A.P. }40,37,34,31,\ldots$
$\displaystyle \hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle a=40,\quad d=37-40=-3$
$\displaystyle \text{We know, }a_n=a+(n-1)d.$
$\displaystyle 0=40+(n-1)(-3)$
$\displaystyle 0=40-3n+3\Rightarrow3n=43\Rightarrow n=\frac{43}{3}$
$\displaystyle \text{Since }n\text{ is not a natural number, }0\text{ is not a term of the A.P.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{If the seventh term of an A.P. is }\frac19\text{ and its ninth term is }\frac17,\text{ find its }63\text{rd term.}$
$\displaystyle \hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know, }a_n=a+(n-1)d.$
$\displaystyle a_7=a+6d=\frac19\qquad\ldots\text{(i)}$
$\displaystyle a_9=a+8d=\frac17\qquad\ldots\text{(ii)}$
$\displaystyle \text{Subtracting (i) from (ii), }2d=\frac17-\frac19=\frac{2}{63}$
$\displaystyle \therefore d=\frac{1}{63}$
$\displaystyle \text{Putting }d=\frac{1}{63}\text{ in (i), }a+\frac{6}{63}=\frac19=\frac{7}{63}$
$\displaystyle \therefore a=\frac{1}{63}$
$\displaystyle a_{63}=a+62d=\frac{1}{63}+\frac{62}{63}=1$
$\displaystyle \therefore \text{The }63\text{rd term is }1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{Which term of the A.P. }3,15,27,39,\ldots\text{ will be }120\text{ more than its }21\text{st term?}$
$\displaystyle \hfill\text{[CBSE 2009]}$
$\displaystyle \text{Answer:}$
$\displaystyle a=3,\quad d=15-3=12$
$\displaystyle \text{We know, }a_n=a+(n-1)d.$
$\displaystyle a_{21}=3+(21-1)(12)=3+240=243$
$\displaystyle \text{Let the }n\text{th term be }120\text{ more than the }21\text{st term.}$
$\displaystyle a_n=243+120=363$
$\displaystyle 363=3+(n-1)(12)$
$\displaystyle 360=12(n-1)\Rightarrow n-1=30\Rightarrow n=31$
$\displaystyle \therefore \text{The }31\text{st term is }120\text{ more than the }21\text{st term.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{For what value of }n,\text{ the }n\text{th terms of the arithmetic progressions }63,65,67,\ldots$
$\displaystyle \text{and }3,10,17,\ldots\text{ are equal?}\hfill\text{[CBSE 2008]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the first A.P., }a=63,\quad d=2$
$\displaystyle a_n=63+(n-1)2=2n+61$
$\displaystyle \text{For the second A.P., }a=3,\quad d=7$
$\displaystyle a_n=3+(n-1)7=7n-4$
$\displaystyle \text{Since their }n\text{th terms are equal, }2n+61=7n-4$
$\displaystyle 5n=65\Rightarrow n=13$
$\displaystyle \therefore \text{The }13\text{th terms of the two A.P.s are equal.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{Find the arithmetic progression whose third term is }$
$\displaystyle 16\text{ and seventh term exceeds} \ \text{its fifth term by }12.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be the first term and }d\text{ be the common difference.}$
$\displaystyle a_3=a+2d=16\qquad\ldots\text{(i)}$
$\displaystyle a_7-a_5=12$
$\displaystyle (a+6d)-(a+4d)=12$
$\displaystyle 2d=12\Rightarrow d=6$
$\displaystyle \text{Putting }d=6\text{ in (i), }a+12=16\Rightarrow a=4$
$\displaystyle \therefore \text{The required A.P. is }4,10,16,22,\ldots$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{The sum of }4\text{th and }8\text{th terms of an A.P. is }24\text{ and the sum of }6\text{th and}$
$\displaystyle 10\text{th terms is }34.\text{ Find the first term and the common difference of the A.P.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be the first term and }d\text{ be the common difference.}$
$\displaystyle a_4+a_8=24$
$\displaystyle (a+3d)+(a+7d)=24$
$\displaystyle 2a+10d=24\Rightarrow a+5d=12\qquad\ldots\text{(i)}$
$\displaystyle a_6+a_{10}=34$
$\displaystyle (a+5d)+(a+9d)=34$
$\displaystyle 2a+14d=34\Rightarrow a+7d=17\qquad\ldots\text{(ii)}$
$\displaystyle \text{Subtracting (i) from (ii), }2d=5\Rightarrow d=\frac52$
$\displaystyle \text{Putting }d=\frac52\text{ in (i), }a+\frac{25}{2}=12$
$\displaystyle a=\frac{24}{2}-\frac{25}{2}=-\frac12$
$\displaystyle \therefore \text{The first term is }-\frac12\text{ and the common difference is }\frac52.$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{The sum of the }5\text{th and the }7\text{th terms of an A.P. is }52\text{ and the }10\text{th term is }46.$
$\displaystyle \text{Find the A.P.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be the first term and }d\text{ be the common difference.}$
$\displaystyle a_5+a_7=52$
$\displaystyle (a+4d)+(a+6d)=52$
$\displaystyle 2a+10d=52\Rightarrow a+5d=26\qquad\ldots\text{(i)}$
$\displaystyle a_{10}=a+9d=46\qquad\ldots\text{(ii)}$
$\displaystyle \text{Subtracting (i) from (ii), }4d=20\Rightarrow d=5$
$\displaystyle \text{Putting }d=5\text{ in (i), }a+25=26\Rightarrow a=1$
$\displaystyle \therefore \text{The required A.P. is }1,6,11,16,\ldots$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{If the sum of }3\text{rd and }8\text{th terms of an A.P. is }7\text{ and the sum of the }7\text{th and}$
$\displaystyle 14\text{th terms is }-3,\text{ find the }10\text{th term.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be the first term and }d\text{ be the common difference.}$
$\displaystyle a_3+a_8=7$
$\displaystyle (a+2d)+(a+7d)=7$
$\displaystyle 2a+9d=7\qquad\ldots\text{(i)}$
$\displaystyle a_7+a_{14}=-3$
$\displaystyle (a+6d)+(a+13d)=-3$
$\displaystyle 2a+19d=-3\qquad\ldots\text{(ii)}$
$\displaystyle \text{Subtracting (i) from (ii), }10d=-10\Rightarrow d=-1$
$\displaystyle \text{Putting }d=-1\text{ in (i), }2a-9=7\Rightarrow a=8$
$\displaystyle a_{10}=a+9d=8+9(-1)=-1$
$\displaystyle \therefore \text{The }10\text{th term is }-1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{If }10\text{ times the }10\text{th term of an A.P. is equal to }15\text{ times the }15\text{th term, show}$
$\displaystyle \text{that }25\text{th term of the A.P. is zero.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be the first term and }d\text{ be the common difference.}$
$\displaystyle \text{Given, }10a_{10}=15a_{15}$
$\displaystyle 10(a+9d)=15(a+14d)$
$\displaystyle 10a+90d=15a+210d$
$\displaystyle 5a+120d=0$
$\displaystyle a+24d=0$
$\displaystyle \text{But }a_{25}=a+(25-1)d=a+24d$
$\displaystyle \therefore a_{25}=0.$
$\displaystyle \text{Hence, the }25\text{th term of the A.P. is zero.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{If }9\text{th term of an A.P. is zero, prove that its }29\text{th term is double the }19\text{th term.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be the first term and }d\text{ be the common difference.}$
$\displaystyle a_9=a+8d=0\qquad\ldots\text{(i)}$
$\displaystyle a_{19}=a+18d=(a+8d)+10d=10d$
$\displaystyle a_{29}=a+28d=(a+8d)+20d=20d$
$\displaystyle \therefore a_{29}=2(10d)=2a_{19}$
$\displaystyle \text{Hence, the }29\text{th term is double the }19\text{th term.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{In a certain A.P. the }24\text{th term is twice the }10\text{th term. Prove that the }72\text{nd}$
$\displaystyle \text{term is twice the }34\text{th term.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be the first term and }d\text{ be the common difference.}$
$\displaystyle \text{Given, }a_{24}=2a_{10}$
$\displaystyle a+23d=2(a+9d)$
$\displaystyle a+23d=2a+18d\Rightarrow a=5d\qquad\ldots\text{(i)}$
$\displaystyle a_{72}=a+71d=5d+71d=76d$
$\displaystyle a_{34}=a+33d=5d+33d=38d$
$\displaystyle \therefore a_{72}=76d=2(38d)=2a_{34}$
$\displaystyle \text{Hence, the }72\text{nd term is twice the }34\text{th term.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{The }26\text{th, }11\text{th and last term of an A.P. are }0,3\text{ and }-\frac15,\text{ respectively. Find}$
$\displaystyle \text{the common difference and the number of terms.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be the first term and }d\text{ be the common difference.}$
$\displaystyle a_{26}=a+25d=0\qquad\ldots\text{(i)}$
$\displaystyle a_{11}=a+10d=3\qquad\ldots\text{(ii)}$
$\displaystyle \text{Subtracting (ii) from (i), }15d=-3\Rightarrow d=-\frac15$
$\displaystyle \text{Putting }d=-\frac15\text{ in (ii), }a-2=3\Rightarrow a=5$
$\displaystyle \text{Since the last term is }-\frac15,\quad-\frac15=5+(n-1)\left(-\frac15\right)$
$\displaystyle -1=25-(n-1)\Rightarrow n-1=26\Rightarrow n=27$
$\displaystyle \therefore \text{The common difference is }-\frac15\text{ and the A.P. has }27\text{ terms.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{The }4\text{th term of an A.P. is three times the first and the }7\text{th term exceeds twice}$
$\displaystyle \text{the third term by }1.\text{ Find the first term and the common difference.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be the first term and }d\text{ be the common difference.}$
$\displaystyle a_4=3a_1$
$\displaystyle a+3d=3a\Rightarrow 3d=2a\qquad\ldots\text{(i)}$
$\displaystyle a_7=2a_3+1$
$\displaystyle a+6d=2(a+2d)+1$
$\displaystyle 2d-a=1\qquad\ldots\text{(ii)}$
$\displaystyle \text{From (i), }a=\frac{3d}{2}$
$\displaystyle \text{Putting this in (ii), }2d-\frac{3d}{2}=1\Rightarrow\frac{d}{2}=1\Rightarrow d=2$
$\displaystyle \text{Therefore, }a=\frac{3(2)}{2}=3$
$\displaystyle \therefore \text{The first term is }3\text{ and the common difference is }2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{Write the expression }a_n-a_k\text{ for the A.P. }a,a+d,a+2d,\ldots$
$\displaystyle \text{Hence, find the common difference of the A.P. for which}$
$\displaystyle \text{(i) }11\text{th term is }5\text{ and }13\text{th term is }79\qquad\text{(ii) }a_{10}-a_5=200$
$\displaystyle \text{(iii) }20\text{th term is }10\text{ more than the }18\text{th term.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know, }a_n=a+(n-1)d\text{ and }a_k=a+(k-1)d.$
$\displaystyle a_n-a_k=[a+(n-1)d]-[a+(k-1)d]$
$\displaystyle \therefore a_n-a_k=(n-k)d.$

$\displaystyle \text{(i) }a_{13}-a_{11}=(13-11)d$
$\displaystyle 79-5=2d$
$\displaystyle 74=2d\Rightarrow d=37$
$\displaystyle \therefore \text{The common difference is }37.$

$\displaystyle \text{(ii) }a_{10}-a_5=(10-5)d$
$\displaystyle 200=5d\Rightarrow d=40$
$\displaystyle \therefore \text{The common difference is }40.$

$\displaystyle \text{(iii) }a_{20}-a_{18}=10$
$\displaystyle (20-18)d=10$
$\displaystyle 2d=10\Rightarrow d=5$
$\displaystyle \therefore \text{The common difference is }5.$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\text{The eighth term of an A.P. is half of its second term and the eleventh}$
$\displaystyle \text{term exceeds one third of its fourth term by }1.\text{ Find the }15\text{th term.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be the first term and }d\text{ be the common difference.}$
$\displaystyle a_8=\frac12a_2$
$\displaystyle a+7d=\frac12(a+d)$
$\displaystyle 2a+14d=a+d\Rightarrow a+13d=0\qquad\ldots\text{(i)}$
$\displaystyle a_{11}=\frac13a_4+1$
$\displaystyle a+10d=\frac13(a+3d)+1$
$\displaystyle 3a+30d=a+3d+3\Rightarrow2a+27d=3\qquad\ldots\text{(ii)}$
$\displaystyle \text{From (i), }a=-13d$
$\displaystyle \text{Putting this in (ii), }-26d+27d=3\Rightarrow d=3$
$\displaystyle \therefore a=-13(3)=-39$
$\displaystyle a_{15}=a+14d=-39+14(3)=3$
$\displaystyle \therefore \text{The }15\text{th term is }3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }\text{Two arithmetic progressions have the same common difference.}$
$\displaystyle \text{The difference between their }100\text{th terms is }100;\text{ what is the difference} \\ \text{between their }1000\text{th terms?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the first terms of the two A.P.s be }a\text{ and }b,\text{ and their common difference be }d.$
$\displaystyle a_{100}=a+99d,\quad b_{100}=b+99d$
$\displaystyle a_{100}-b_{100}=(a+99d)-(b+99d)=a-b=100\qquad\ldots\text{(i)}$
$\displaystyle a_{1000}=a+999d,\quad b_{1000}=b+999d$
$\displaystyle a_{1000}-b_{1000}=(a+999d)-(b+999d)=a-b$
$\displaystyle \text{From (i), }a-b=100$
$\displaystyle \therefore \text{The difference between their }1000\text{th terms is }100.$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }\text{How many multiples of }4\text{ lie between }10\text{ and }250?$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The multiples of }4\text{ between }10\text{ and }250\text{ are }12,16,20,\ldots,248.$
$\displaystyle \text{Here, }a=12,\quad d=4,\quad a_n=248$
$\displaystyle \text{We know, }a_n=a+(n-1)d.$
$\displaystyle 248=12+(n-1)4$
$\displaystyle 236=4(n-1)\Rightarrow n-1=59\Rightarrow n=60$
$\displaystyle \therefore \text{There are }60\text{ multiples of }4\text{ between }10\text{ and }250.$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: }\text{How many three digit numbers are divisible by }7?\hfill\text{[CBSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The first three digit number divisible by }7\text{ is }105.$
$\displaystyle \text{The last three digit number divisible by }7\text{ is }994.$
$\displaystyle \text{Thus, the numbers are }105,112,119,\ldots,994.$
$\displaystyle \text{Here, }a=105,\quad d=7,\quad a_n=994$
$\displaystyle \text{We know, }a_n=a+(n-1)d.$
$\displaystyle 994=105+(n-1)7$
$\displaystyle 889=7(n-1)\Rightarrow n-1=127\Rightarrow n=128$
$\displaystyle \therefore \text{There are }128\text{ three digit numbers divisible by }7.$
$\displaystyle \\$

$\displaystyle \textbf{Question 27: }\text{The }17\text{th term of an A.P. is }5\text{ more than twice its }8\text{th term. If the }11\text{th term}$
$\displaystyle \text{of the A.P. is }43,\text{ find the }n\text{th term.}\hfill\text{[CBSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be the first term and }d\text{ be the common difference.}$
$\displaystyle a_{17}=2a_8+5$
$\displaystyle a+16d=2(a+7d)+5$
$\displaystyle a+16d=2a+14d+5\Rightarrow2d-a=5\qquad\ldots\text{(i)}$
$\displaystyle a_{11}=a+10d=43\qquad\ldots\text{(ii)}$
$\displaystyle \text{From (i), }a=2d-5$
$\displaystyle \text{Putting this in (ii), }2d-5+10d=43$
$\displaystyle 12d=48\Rightarrow d=4$
$\displaystyle \therefore a=2(4)-5=3$
$\displaystyle a_n=a+(n-1)d=3+4(n-1)=4n-1$
$\displaystyle \therefore \text{The }n\text{th term is }4n-1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 28: }\text{Find the number of all three digit natural numbers which are divisible by }9.$
$\displaystyle \hfill\text{[CBSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The first three digit natural number divisible by }9\text{ is }108.$
$\displaystyle \text{The last three digit natural number divisible by }9\text{ is }999.$
$\displaystyle \text{Thus, the numbers are }108,117,126,\ldots,999.$
$\displaystyle \text{Here, }a=108,\quad d=9,\quad a_n=999$
$\displaystyle \text{We know, }a_n=a+(n-1)d.$
$\displaystyle 999=108+(n-1)9$
$\displaystyle 891=9(n-1)\Rightarrow n-1=99\Rightarrow n=100$
$\displaystyle \therefore \text{There are }100\text{ three digit natural numbers divisible by }9.$
$\displaystyle \\$

$\displaystyle \textbf{Question 29: }\text{The }19\text{th term of an A.P. is equal to three times its sixth term.}$
$\displaystyle \text{If its }9\text{th term} \ \text{is }19,\text{ find the A.P.}\hfill\text{[CBSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be the first term and }d\text{ be the common difference.}$
$\displaystyle a_{19}=3a_6$
$\displaystyle a+18d=3(a+5d)$
$\displaystyle a+18d=3a+15d\Rightarrow3d=2a\qquad\ldots\text{(i)}$
$\displaystyle a_9=a+8d=19\qquad\ldots\text{(ii)}$
$\displaystyle \text{From (i), }a=\frac{3d}{2}$
$\displaystyle \text{Putting this in (ii), }\frac{3d}{2}+8d=19$
$\displaystyle \frac{19d}{2}=19\Rightarrow d=2$
$\displaystyle \therefore a=\frac{3(2)}{2}=3$
$\displaystyle \therefore \text{The required A.P. is }3,5,7,9,\ldots$
$\displaystyle \\$

$\displaystyle \textbf{Question 30: }\text{Find the number of natural numbers between }101\text{ and }999$
$\displaystyle \text{ which are divisible} \ \text{by both }2\text{ and }5.\hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{A number divisible by both }2\text{ and }5\text{ is divisible by }10.$
$\displaystyle \text{The required numbers are }110,120,130,\ldots,990.$
$\displaystyle \text{Here, }a=110,\quad d=10,\quad a_n=990$
$\displaystyle \text{We know, }a_n=a+(n-1)d.$
$\displaystyle 990=110+(n-1)10$
$\displaystyle 880=10(n-1)\Rightarrow n-1=88\Rightarrow n=89$
$\displaystyle \therefore \text{There are }89\text{ natural numbers divisible by both }2\text{ and }5.$
$\displaystyle \\$

$\displaystyle \textbf{Question 31: }\text{Find the middle term of the A.P. }213,205,197,\ldots,37.\hfill\text{[CBSE 2015]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Here, }a=213,\quad d=205-213=-8,\quad a_n=37$
$\displaystyle \text{We know, }a_n=a+(n-1)d.$
$\displaystyle 37=213+(n-1)(-8)$
$\displaystyle -176=-8(n-1)\Rightarrow n-1=22\Rightarrow n=23$
$\displaystyle \text{Since }n=23\text{ is odd, the middle term is the }\left(\frac{23+1}{2}\right)\text{th}=12\text{th term.}$
$\displaystyle a_{12}=213+(12-1)(-8)=213-88=125$
$\displaystyle \therefore \text{The middle term is }125.$
$\displaystyle \\$

$\displaystyle \textbf{Question 32: }\text{Find the sum of two middle terms of the A.P. }-\frac43,-1,-\frac23,-\frac13,\ldots,4\frac13.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Here, }a=-\frac43,\quad d=-1-\left(-\frac43\right)=\frac13,\quad a_n=4\frac13=\frac{13}{3}$
$\displaystyle \text{We know, }a_n=a+(n-1)d.$
$\displaystyle \frac{13}{3}=-\frac43+(n-1)\frac13$
$\displaystyle 13=-4+n-1\Rightarrow n=18$
$\displaystyle \text{Since }n=18\text{ is even, the two middle terms are the }9\text{th and }10\text{th terms.}$
$\displaystyle a_9=-\frac43+(9-1)\frac13=\frac43$
$\displaystyle a_{10}=-\frac43+(10-1)\frac13=\frac53$
$\displaystyle a_9+a_{10}=\frac43+\frac53=\frac93=3$
$\displaystyle \therefore \text{The sum of the two middle terms is }3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 33: }\text{A sum of Rs. 2,000 is invested at }7\%\text{ per annum simple}$
$\displaystyle \text{interest. Calculate the interests at the end of }1\text{st, }2\text{nd and }3\text{rd year. Do these }$
$\displaystyle \text{interests form an A.P.? If so, find the interest at the end of the }27\text{th year.}\hfill\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Here, }P=\text{Rs. }2000,\quad R=7\%\text{ per annum.}$
$\displaystyle \text{Simple interest after }n\text{ years}=\frac{PRn}{100}$
$\displaystyle I_n=\frac{2000\times7\times n}{100}=140n$
$\displaystyle I_1=\text{Rs. }140,\quad I_2=\text{Rs. }280,\quad I_3=\text{Rs. }420$
$\displaystyle 280-140=140,\quad420-280=140$
$\displaystyle \therefore \text{The interests form an A.P. with common difference Rs. }140.$
$\displaystyle I_{27}=140(27)=\text{Rs. }3780$
$\displaystyle \therefore \text{The interest at the end of the }27\text{th year is Rs. }3780.$
$\displaystyle \\$

$\displaystyle \textbf{Question 34: }\text{If }(m+1)\text{th term of an A.P. is twice the }(n+1)\text{th term, prove that }$
$\displaystyle (3m+1)\text{th} \ \text{term is twice the }(m+n+1)\text{th term.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a\text{ be the first term and }d\text{ be the common difference.}$
$\displaystyle \text{Given, }a_{m+1}=2a_{n+1}$
$\displaystyle a+md=2[a+nd]$
$\displaystyle a=(m-2n)d\qquad\ldots\text{(i)}$
$\displaystyle a_{3m+1}=a+3md$
$\displaystyle 2a_{m+n+1}=2[a+(m+n)d]$
$\displaystyle =2a+2md+2nd$
$\displaystyle \text{Using (i), }2a+2md+2nd=a+3md$
$\displaystyle \therefore a_{3m+1}=2a_{m+n+1}$
$\displaystyle \text{Hence, the }(3m+1)\text{th term is twice the }(m+n+1)\text{th term.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 35: }\text{If an A.P. consists of }n\text{ terms with first term }a\text{ and }$
$\displaystyle n\text{th term }l,\text{ show that the sum} \ \text{of the }m\text{th term from the beginning and the } \\ m\text{th term from the end is }(a+l).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }d\text{ be the common difference of the A.P.}$
$\displaystyle m\text{th term from the beginning}=a+(m-1)d$
$\displaystyle m\text{th term from the end}=l-(m-1)d$
$\displaystyle \text{Their sum}=a+(m-1)d+l-(m-1)d$
$\displaystyle =a+l$
$\displaystyle \therefore \text{The required sum is }a+l.$
$\displaystyle \\$

$\displaystyle \textbf{Question 36: }\text{How many numbers lie between }10\text{ and }300,\text{ which when divided by }4\text{ leave}$
$\displaystyle \text{a remainder }3?$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The required numbers are }11,15,19,\ldots,299.$
$\displaystyle \text{Here, }a=11,\quad d=4,\quad a_n=299$
$\displaystyle \text{We know, }a_n=a+(n-1)d.$
$\displaystyle 299=11+(n-1)4$
$\displaystyle 288=4(n-1)\Rightarrow n-1=72\Rightarrow n=73$
$\displaystyle \therefore \text{There are }73\text{ such numbers between }10\text{ and }300.$
$\displaystyle \\$

$\displaystyle \textbf{Question 37: }\text{For the A.P. }-3,-7,-11,\ldots,\text{ can we find }a_{30}-a_{20}\text{ without actually finding}$
$\displaystyle a_{30}\text{ and }a_{20}?\text{ Give reasons for your answer.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Yes. Here, }d=-7-(-3)=-4.$
$\displaystyle \text{We know, }a_n-a_k=(n-k)d.$
$\displaystyle a_{30}-a_{20}=(30-20)(-4)=10(-4)=-40$
$\displaystyle \therefore \text{We can find }a_{30}-a_{20}\text{ directly, and its value is }-40.$
$\displaystyle \\$

$\displaystyle \textbf{Question 38: }\text{Two A.P.s have the same common difference. The first term of}$
$\displaystyle \text{one A.P. is }2\text{ and that} \ \text{of the other is }7.\text{ The difference between their }10\text{th terms}$
$\displaystyle \text{is the same as the difference between their }21\text{st terms, which is the same as}$
$\displaystyle \text{the difference between any two corresponding terms. Why?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the common difference of the two A.P.s be }d.$
$\displaystyle \text{Their }n\text{th terms are }a_n=2+(n-1)d\text{ and }b_n=7+(n-1)d.$
$\displaystyle b_n-a_n=[7+(n-1)d]-[2+(n-1)d]=5$
$\displaystyle \therefore \text{The difference between any two corresponding terms is always }5.$
$\displaystyle \text{Hence, the difference between their }10\text{th, }21\text{st or any corresponding terms is the same.}$
$\displaystyle \\$