\displaystyle \textbf{Question 1: }\text{Determine the general term of an AP whose }7^\text{th}\text{ term is }-1 \\ \text{and }16^\text{th}\text{ term is }17. \ \text{[ICSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }a\text{ be the first term and }d\text{ the common difference.}
\displaystyle a_7=-1\text{ and }a_{16}=17.
\displaystyle a_7=a+(7-1)d=a+6d=-1\quad (i).
\displaystyle a_{16}=a+(16-1)d=a+15d=17\quad (ii).
\displaystyle \text{Subtracting (i) from (ii), }a+15d-a-6d=17-(-1).
\displaystyle 9d=18.
\displaystyle d=2.
\displaystyle \text{Substituting }d=2\text{ in (i), }a+12=-1.
\displaystyle a=-13.
\displaystyle \text{General term }a_n=a+(n-1)d.
\displaystyle a_n=-13+(n-1)2.
\displaystyle a_n=-13+2n-2=2n-15.

\displaystyle \textbf{Question 2: }\text{If }(k-3),(2k+1)\text{ and }(4k+3)\text{ are three consecutive} \\ \text{terms of an AP, find }k.\ \text{[ICSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }a_1=k-3,\ a_2=2k+1\text{ and }a_3=4k+3.
\displaystyle \text{If they are consecutive terms of an AP, then }a_2-a_1=a_3-a_2.
\displaystyle (2k+1)-(k-3)=(4k+3)-(2k+1).
\displaystyle 2k+1-k+3=4k+3-2k-1.
\displaystyle k+4=2k+2.
\displaystyle k=2.

\displaystyle \textbf{Question 3:}\text{The sum of three numbers in AP is }15\text{ and their product is }105. \\ \text{Find the numbers.} \ \text{[ICSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{Let three numbers in AP be }a-d,\ a,\ a+d.
\displaystyle \text{According to the question, sum }=15.
\displaystyle (a-d)+a+(a+d)=15.
\displaystyle 3a=15.
\displaystyle a=5.
\displaystyle \text{Product of numbers }=105.
\displaystyle (a-d)a(a+d)=105.
\displaystyle (5-d)5(5+d)=105.
\displaystyle 5(25-d^2)=105.
\displaystyle 25-d^2=21.
\displaystyle d^2=4.
\displaystyle d=\pm 2.
\displaystyle \text{If }d=2,\text{ numbers are }3,5,7.
\displaystyle \text{If }d=-2,\text{ numbers are }7,5,3.

\displaystyle \textbf{Question 4: }\text{The sum of the first three terms of an AP is }42\text{ and the product of the} \\ \text{first and third term is }52.  \text{Find the first term and the common difference.} \hspace{0.5cm} \text{[ICSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }a\text{ and }d\text{ be the first term and common difference of an AP.}
\displaystyle \text{First three terms are }a,a+d,a+2d.
\displaystyle a+(a+d)+(a+2d)=42.
\displaystyle 3a+3d=42.
\displaystyle a+d=14.
\displaystyle a(a+2d)=52.
\displaystyle a(a+2(14-a))=52.
\displaystyle a(28-a)=52.
\displaystyle 28a-a^2=52.
\displaystyle a^2-28a+52=0.
\displaystyle (a-2)(a-26)=0.
\displaystyle a=2\text{ or }a=26.
\displaystyle \text{When }a=2,\ d=14-2=12.
\displaystyle \text{When }a=26,\ d=14-26=-12.

\displaystyle \textbf{Question 5: }\text{Which term of the AP }21,18,15,\ldots\text{ is }-81\text{?} \hspace{3.0cm} \text{[ICSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, AP is }21,18,15,\ldots.
\displaystyle \text{Here, }a=21\text{ and }d=18-21=-3.
\displaystyle \text{Let the }n\text{th term of the given AP be }-81.
\displaystyle a_n=a+(n-1)d.
\displaystyle -81=21+(n-1)(-3).
\displaystyle -81=21-3n+3.
\displaystyle -81=24-3n.
\displaystyle -3n=-81-24=-105.
\displaystyle n=\frac{-105}{-3}=35.
\displaystyle \text{Hence, 35th term of the given AP is }-81.

\displaystyle \textbf{Question 6: }\text{Determine the }10^\text{th}\text{ term from the end} \text{of the AP }4,9,14,\ldots,254. \\ \text{[ICSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, AP is }4,9,14,\ldots,254.
\displaystyle \text{Here, }l=254\text{ and }d=9-4=5.
\displaystyle \text{10th term from the end }=l-(n-1)d.
\displaystyle =254-9\times 5.
\displaystyle =254-45=209.

\displaystyle \textbf{Question 7: }\text{How many terms are there in the sequence }3,6,9,12,\ldots,111\text{?} \ \text{[ICSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, sequence is }3,6,9,12,\ldots,111.
\displaystyle \text{Here, }6-3=9-6=12-9=3.
\displaystyle \text{So, it is an AP with }a=3\text{ and }d=3.
\displaystyle a_n=a+(n-1)d.
\displaystyle 111=3+(n-1)3.
\displaystyle 111=3n.
\displaystyle n=\frac{111}{3}=37.
\displaystyle \text{Hence, the given sequence contains }37\text{ terms.}

\displaystyle \textbf{Question 8: }\text{The first four terms of an AP, whose first term is }4\text{ and common} \\ \text{difference is }-6,\text{ are}\
\displaystyle \text{(a) }4,-10,-16,-22\qquad\text{(b) }4,10,16,22\qquad\text{(c) }4,-2,-8,-14\qquad\text{(d) }4,2,8,14 \hspace{0.5cm} \text{[ICSE 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given, }a=4\text{ and }d=-6.
\displaystyle \text{First four terms of an AP are }a,a+d,a+2d,a+3d.
\displaystyle =4,4-6,4-2(6),4-3(6).
\displaystyle =4,-2,-8,-14.

\displaystyle \textbf{Question 9: }\text{If }70,75,80\text{ and }85\text{ are the first four terms of an AP, then the }10^\text{th}\text{ term is} \\
\displaystyle \text{(a) }35\qquad\text{(b) }25\qquad\text{(c) }115\qquad\text{(d) }105 \hspace{6.0cm} \text{[ICSE 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given, first four terms of an AP are }70,75,80,85.
\displaystyle \text{Here, }a=70\text{ and }d=75-70=5.
\displaystyle T_n=a+(n-1)d.
\displaystyle T_{10}=70+(10-1)5.
\displaystyle T_{10}=70+45=115.

\displaystyle \textbf{Question 10: }\text{In an AP, first term is }5,\text{ common difference is }-3\text{ and the }n^\text{th}\text{ term is }-7,\text{ then }n\text{ equals}\ 
\displaystyle \text{(a) }5\qquad\text{(b) }17\qquad\text{(c) }-13\qquad\text{(d) }7 \hspace{7.0cm} \text{[ICSE 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, }a=5,d=-3\text{ and }T_n=-7.
\displaystyle T_n=a+(n-1)d.
\displaystyle -7=5+(n-1)(-3).
\displaystyle -7=5-3n+3.
\displaystyle -7=8-3n.
\displaystyle 3n=15.
\displaystyle n=5.

\displaystyle \textbf{Question 11: }\text{The first three terms of an AP are }1,9,17,\text{ then the next two terms are}
\displaystyle \text{(a) }25\text{ and }35\qquad\text{(b) }27\text{ and }37\qquad\text{(c) }25\text{ and }33\qquad\text{(d) None of these} \hspace{3.0cm} \text{[ICSE 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{(c) First term }(a_1)=1\text{ and second term }(a_2)=9.
\displaystyle \text{So, common difference }d=a_2-a_1=9-1=8.
\displaystyle \text{Therefore, next terms are }17+8=25\text{ and }25+8=33.
\displaystyle \text{Hence, next two terms are }25\text{ and }33.

\displaystyle \textbf{Question 12: }\text{If }73\text{ is the }n^\text{th}\text{ term of the AP }3,8,13,18,\ldots\text{ then }n\text{ is}\ 
\displaystyle \text{(a) }13\qquad\text{(b) }14\qquad\text{(c) }15\qquad\text{(d) }16 \hspace{7.0cm} \text{[ICSE 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given, AP is }3,8,13,18,\ldots.
\displaystyle \text{So, }a=3\text{ and }d=8-3=5.
\displaystyle a_n=a+(n-1)d.
\displaystyle 73=3+(n-1)5.
\displaystyle 73=3+5n-5.
\displaystyle 73=5n-2.
\displaystyle 5n=75.
\displaystyle n=15.

\displaystyle \textbf{Question 13: }\text{The }n^\text{th}\text{ term of an AP is }2n+5.\text{ The }10^\text{th}\text{ term is}
\displaystyle \text{(a) }7\qquad\text{(b) }15\qquad\text{(c) }25\qquad\text{(d) }45 \hspace{7.0cm} \text{[ICSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given, the }n\text{th term of an Arithmetic Progression }t_n=2n+5.
\displaystyle \text{For 10th term, put }n=10.
\displaystyle t_{10}=2(10)+5=20+5.
\displaystyle t_{10}=25.
\displaystyle \text{Hence, option (c) is correct.}

\displaystyle \textbf{Question 14: }\text{ }57,54,51,48,\ldots\text{ are in AP. The value of the }8^\text{th}\text{ term is}
\displaystyle \text{(a) }36\qquad\text{(b) }78\qquad\text{(c) }-36\qquad\text{(d) }-78 \hspace{7.0cm} \text{[ICSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, AP is }57,54,51,48.
\displaystyle \text{Here, }a=57\text{ and }d=54-57=-3.
\displaystyle \text{The }n\text{th term of AP is }a_n=a+(n-1)d.
\displaystyle a_8=57+(8-1)(-3).
\displaystyle a_8=57-21=36.
\displaystyle \text{Hence, 8th term of AP is }36.

\displaystyle \textbf{Question 15: }\text{If the }n^\text{th}\text{ term of an AP is }(n+3),\text{ then the first three terms are}
\displaystyle \text{(a) }1,2,3\qquad\text{(b) }2,4,6\qquad\text{(c) }4,5,6\qquad\text{(d) }7,8,9 \hspace{6.0cm} \text{[ICSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given, the }n\text{th term of an AP is }(n+3).
\displaystyle \text{i.e., }a_n=n+3.
\displaystyle \text{On putting }n=1,\ a_1=1+3=4.
\displaystyle \text{On putting }n=2,\ a_2=2+3=5.
\displaystyle \text{On putting }n=3,\ a_3=3+3=6.
\displaystyle \text{Hence, the first three terms of the AP are }4,5,6.

\displaystyle \textbf{Question 17: }\text{The }7^\text{th}\text{ term of the AP }\frac{1}{a},\left(\frac{1}{a}+1\right),\left(\frac{1}{a}+2\right),\ldots\text{ is}. \hspace{1.0cm} \text{[ICSE 2024]}
\displaystyle \text{(a) }\left(\frac{1}{a}+6\right)\qquad\text{(b) }\left(\frac{1}{a}+7\right)\qquad\text{(c) }\left(\frac{1}{a}+8\right)\qquad\text{(d) }\left(\frac{1}{a}+77\right)
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given AP, }\frac{1}{a},\left(\frac{1}{a}+1\right),\left(\frac{1}{a}+2\right),\ldots
\displaystyle \text{First term }a_1=\frac{1}{a}.
\displaystyle \text{Common difference }d=\left(\frac{1}{a}+1\right)-\frac{1}{a}=1.
\displaystyle a_7=a_1+(7-1)d.
\displaystyle =\frac{1}{a}+6.

\displaystyle \textbf{Question 18: }\text{Which of the following are AP's? If they form an AP, then find the} \\ \text{common difference and write three more terms.} \ \text{[ICSE 2024]}
\displaystyle \text{(i) }2,4,8,16,\ldots
\displaystyle \text{(ii) }2,\frac{5}{2},3,\frac{7}{2},\ldots
\displaystyle \text{(iii) }-1.2,-3.2,-5.2,-7.2,\ldots
\displaystyle \text{Answer:}
\displaystyle \text{i) Here, }a_2-a_1=4-2=2\text{ and }a_3-a_2=8-4=4.
\displaystyle \text{Since }a_2-a_1\ne a_3-a_2,\text{ it is not an AP.}
\displaystyle \text{(ii) }a_2-a_1=\frac{5}{2}-2=\frac{1}{2}.
\displaystyle a_3-a_2=\frac{5}{2}-2=\frac{1}{2}.
\displaystyle \text{Common difference }d=\frac{1}{2}.
\displaystyle a_5=a_4+d=\frac{7}{2}+\frac{1}{2}=4.
\displaystyle a_6=a_5+d=4+\frac{1}{2}=\frac{9}{2}.
\displaystyle a_7=a_6+d=\frac{9}{2}+\frac{1}{2}=5.
\displaystyle \text{(iii) }a_2-a_1=-3.2-(-1.2)=-2.
\displaystyle a_3-a_2=-5.2-(-3.2)=-2.
\displaystyle \text{Common difference }d=-2.
\displaystyle a_5=-7.2+(-2)=-9.2.
\displaystyle a_6=-9.2+(-2)=-11.2.
\displaystyle a_7=-11.2+(-2)=-13.2.

\displaystyle \textbf{Question 19: }\text{The 4th term of an AP is }22\text{ and 15th term is }66.  \text{Find} \\ \text{the first term and the common difference. Hence, find the sum of the series to} \\ \text{ 8 terms. [ICSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }T_4=22.
\displaystyle a+3d=22.\quad (i)
\displaystyle T_{15}=66.
\displaystyle a+14d=66.\quad (ii)
\displaystyle \text{Subtract (i) from (ii): }a+14d-(a+3d)=66-22.
\displaystyle 11d=44.
\displaystyle d=\frac{44}{11}=4.
\displaystyle \text{Substitute }d=4\text{ in (i): }a+3\times4=22.
\displaystyle a+12=22.
\displaystyle a=10.
\displaystyle S_8=\frac{8}{2}[2\times10+(8-1)\times4].
\displaystyle =4[20+28]=4\times48=192.

\displaystyle \textbf{Question 20: }\text{In an AP the 4th and 6th terms are }8\text{ and }14\text{ respectively.}
\displaystyle \text{Find (i) first term (ii) common difference (iii) sum of the first 20 terms. [ICSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let first term }=a\text{ and common difference }=d.
\displaystyle a+3d=8.\quad (i)
\displaystyle a+5d=14.\quad (ii)
\displaystyle \text{Subtract (i) from (ii): }2d=6.
\displaystyle d=3.
\displaystyle a+3\times3=8\Rightarrow a+9=8.
\displaystyle a=-1.
\displaystyle S_{20}=\frac{20}{2}[2(-1)+(20-1)\times3].
\displaystyle =10[-2+57]=10\times55=550.
\displaystyle \text{(i) First term }a=-1.
\displaystyle \text{(ii) Common difference }d=3.
\displaystyle \text{(iii) Sum of first 20 terms.}
\displaystyle S_{20}=\frac{20}{2}[2(-1)+(20-1)3].
\displaystyle =10[-2+57]=10\times55=550.

\displaystyle \textbf{Question 21: }\text{The 2nd and 45th term of an arithmetic progression are }10\text{ and }96 \\ \text{ respectively.} \text{Find the first term and the common difference and hence find the sum of} \\ \text{the first 15 terms. [ICSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Let first term }=a\text{ and common difference }=d.
\displaystyle a_2=a+d=10.\quad (i)
\displaystyle a_{45}=a+44d=96.\quad (ii)
\displaystyle \text{From (i), }a=10-d.
\displaystyle \text{Substitute in (ii): }(10-d)+44d=96.
\displaystyle 10+43d=96.
\displaystyle 43d=86.
\displaystyle d=2.
\displaystyle a=10-2=8.
\displaystyle S_{15}=\frac{15}{2}[2\times8+(15-1)\times2].
\displaystyle =\frac{15}{2}[16+28]=\frac{15}{2}\times44.
\displaystyle =15\times22=330.

\displaystyle \textbf{Question 22: }\text{Find the 16th term of the AP }7,11,15,19,\ldots\text{ and find the} \\ \text{sum of the first 6 terms. [ICSE Specimen 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, AP is }7,11,15,19,\ldots
\displaystyle a=7\text{ and }d=11-7=4.
\displaystyle a_{16}=a+(16-1)d=7+15\times4=7+60=67.
\displaystyle S_6=\frac{6}{2}[2\times7+(6-1)\times4].
\displaystyle =3[14+20]=3\times34=102.

\displaystyle \textbf{Question 23: }\text{If the 6th term of an AP is equal to four times its first term and the sum} \\ \text{of first six terms is }75,  \text{find the first term and the common difference. [ICSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Let first term }=a\text{ and common difference }=d.
\displaystyle T_6=4a.
\displaystyle a+(6-1)d=4a.
\displaystyle a+5d=4a.
\displaystyle 3a=5d.\quad (i)
\displaystyle S_6=75.
\displaystyle \frac{6}{2}[2a+(6-1)d]=75.
\displaystyle 3(2a+5d)=75.
\displaystyle 2a+5d=25.\quad (ii)
\displaystyle \text{From (i) and (ii), }2a+3a=25.
\displaystyle 5a=25.
\displaystyle a=5.
\displaystyle 5d=3a=15.
\displaystyle d=3.

\displaystyle \textbf{Question 24: }\text{The }n\text{th term of an Arithmetic Progression (AP) is }(3n+1).
\displaystyle \text{(i) The first three terms of this AP are }(a)\ 5,6,7\ (b)\ 3,6,9\ (c)\ 1,4,7\ (d)\ 4,7,10.
\displaystyle \text{(ii) The common difference of the AP is }(a)\ 3\ (b)\ 1\ (c)\ -3\ (d)\ 2.
\displaystyle \text{(iii) Which of the following is not a term of this AP? }(a)\ 25\ (b)\ 27\ (c)\ 28\ (d)\ 31.
\displaystyle \text{(iv) Sum of the first 10 terms of this AP is }(a)\ 350\ (b)\ 175\ (c)\ -95\ (d)\ 70.
\displaystyle \text{[ICSE Specimen Semester-I 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{(i ) (d) Given, }a_n=3n+1.
\displaystyle a_1=3\times1+1=4,\ a_2=3\times2+1=7,\ a_3=3\times3+1=10.
\displaystyle \text{First three terms are }4,7,10.
\displaystyle \text{(ii ) (a) } d=a_2-a_1=7-4=3.
\displaystyle \text{(iii ) ( b) Let }a_n=27\Rightarrow 3n+1=27.
\displaystyle 3n=26\Rightarrow n=\frac{26}{3}.
\displaystyle \text{Since }n\text{ is not an integer, 27 is not a term.}
\displaystyle S_{10}=\frac{10}{2}[2\times4+(10-1)\times3].
\displaystyle =5[8+27]=5\times35=175.

\displaystyle \textbf{Question 25: }\text{The sum of the first 20 terms of an AP }2,4,6,8,\ldots\text{ is }
\displaystyle \text{(a) }400\qquad \text{(b) }840\qquad \text{(c) }420\qquad \text{(d) }800\ \text{[ICSE Semester I 2022]}
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given AP is }2,4,6,8,\ldots
\displaystyle a=2\text{ and }d=4-2=2.
\displaystyle S_n=\frac{n}{2}[2a+(n-1)d].
\displaystyle S_{20}=\frac{20}{2}[2\times2+(20-1)\times2].
\displaystyle =10[4+38]=10\times42=420.

\displaystyle \textbf{Question 26: }\text{The first, the last term and the common difference of an AP are } \\ 98,1001\text{ and }7\text{ respectively.}  \text{Find (i) number of terms }n\text{ (ii) sum of }n \\ \text{ terms. [ICSE Specimen 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) } \text{Given, }a_1=98,\ a_{130}=1001\text{ and }d=7.
\displaystyle a_n=a+(n-1)d.
\displaystyle 1001=98+(n-1)7.
\displaystyle 1001-98=7(n-1).
\displaystyle 903=7(n-1).
\displaystyle n-1=129.
\displaystyle n=130.
\displaystyle \text{(ii) } S_{130}=\frac{130}{2}(98+1001).
\displaystyle =65\times1099=71435.

\displaystyle \textbf{Question 27: }\text{The 5th term and the 9th term of an Arithmetic Progression are } \\ 4\text{ and }-12 \text{ respectively.} \text{Find (i) the first term (ii) common difference (iii) sum of 16} \\ \text{terms of the AP. [ICSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a_5=4\text{ and }a_9=-12.
\displaystyle a+4d=4.\quad (i)
\displaystyle a+8d=-12.\quad (ii)
\displaystyle \text{Subtract (i) from (ii): }4d=-16.
\displaystyle d=-4.
\displaystyle a+4(-4)=4\Rightarrow a-16=4.
\displaystyle a=20.
\displaystyle S_{16}=\frac{16}{2}[2\times20+(16-1)(-4)].
\displaystyle =8[40-60]=8\times(-20)=-160.

\displaystyle \textbf{Question 28: }\text{Which term of the Arithmetic Progression (AP) }15,30,45,60,\ldots\text{ is }300?
\displaystyle \text{Hence, find the sum of all the terms of the Arithmetic Progression (AP). [ICSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Formula for }n\text{th term of AP is }t_n=a+(n-1)d.
\displaystyle \text{Here, AP is }15,30,45,60,\ldots\text{ and last term }=300.
\displaystyle a=15\text{ and }d=30-15=15.
\displaystyle 300=15+(n-1)15.
\displaystyle 300=15+15n-15.
\displaystyle 300=15n.
\displaystyle n=20.
\displaystyle S_n=\frac{n}{2}(a+t_n).
\displaystyle S_{20}=\frac{20}{2}(15+300)=10\times315=3150.

\displaystyle \textbf{Question 29: }\text{The }n\text{th term of an Arithmetic Progression (AP) is given by }T_n=6(7-n). \text{ Find}
\displaystyle \text{(i) its first term and common difference.}
\displaystyle \text{(ii) sum of its first 25 terms. [ICSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }n\text{th term of AP is }T_n=6(7-n).
\displaystyle \text{(i) For first term, put }n=1.
\displaystyle T_1=6(7-1)=6\times6=36.
\displaystyle \text{For second term, }T_2=6(7-2)=6\times5=30.
\displaystyle d=T_2-T_1=30-36=-6.
\displaystyle \text{(ii) Sum of first 25 terms.}
\displaystyle S_{25}=\frac{25}{2}[2\times36+(25-1)\times(-6)].
\displaystyle =\frac{25}{2}[72-144]=\frac{25}{2}\times(-72).
\displaystyle =25\times(-36)=-900.

\displaystyle \textbf{Question 30: }\text{The 4th term of a GP is }16\text{ and 7th term is }128.
\displaystyle \text{Find the first term and common ratio of the series. [ICSE 2018]}
\displaystyle \text{Answer:}

\displaystyle \text{Given, }T_4=16\text{ and }T_7=128.
\displaystyle T_4=ar^{4-1}=ar^3=16.\quad (i)
\displaystyle T_7=ar^{7-1}=ar^6=128.\quad (ii)
\displaystyle \text{From (i) and (ii), }\frac{ar^6}{ar^3}=\frac{128}{16}.
\displaystyle r^3=8.
\displaystyle r=2.
\displaystyle \text{Put }r=2\text{ in (i): }ar^3=16\Rightarrow a(2)^3=16.
\displaystyle 8a=16\Rightarrow a=2.
\displaystyle \text{Hence, the first term is }2\text{ and common ratio is }2.

\displaystyle \textbf{Question 31: }\text{15, 30, 60, 120, }\ldots\text{ are in GP (Geometric Progression).}
\displaystyle \text{(i) Find the }n\text{th term of this GP in terms of }n.
\displaystyle \text{(ii) How many terms of the above GP will give the sum }\text{Rs }945\ ?\ \text{[ICSE 2024]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }15,30,60,120,\ldots\text{ are in GP.}
\displaystyle \text{(i) We have, first term }(a)=15.
\displaystyle \text{Common ratio }(r)=\frac{30}{15}=2.
\displaystyle \therefore\ \text{n term }a_n=ar^{\,n-1}.
\displaystyle =15\cdot2^{\,n-1}=\frac{15}{2}\cdot2^n=7.5\cdot2^n.
\displaystyle \text{(ii) Let sum of }n\text{ terms give the sum }945.
\displaystyle \text{We know that }S_n=\frac{a(r^n-1)}{r-1},\ r>1.
\displaystyle \Rightarrow 945=\frac{15(2^n-1)}{2-1}.
\displaystyle \Rightarrow 945=15(2^n-1).
\displaystyle \Rightarrow 63=2^n-1.
\displaystyle \Rightarrow 2^n=64=2^6\Rightarrow n=6.

\displaystyle \textbf{Question 32: }\text{If the sum of first 10 terms of an AP is }140\text{ and the sum of first 16 terms is }320,
\displaystyle \text{then find the sum of first }m\text{ terms.}
\displaystyle \text{Answer:}
\displaystyle \text{Let first term }=a\text{ and common difference }=d.
\displaystyle S_{10}=140.
\displaystyle \frac{10}{2}[2a+(10-1)d]=140.
\displaystyle 5(2a+9d)=140.
\displaystyle 2a+9d=28.\quad (i)
\displaystyle S_{16}=320.
\displaystyle \frac{16}{2}[2a+(16-1)d]=320.
\displaystyle 8(2a+15d)=320.
\displaystyle 2a+15d=40.\quad (ii)
\displaystyle \text{Subtract (i) from (ii): }2a+15d-2a-9d=40-28.
\displaystyle 6d=12.
\displaystyle d=2.
\displaystyle \text{Substitute }d=2\text{ in (i): }2a+9\times2=28.
\displaystyle 2a+18=28.
\displaystyle 2a=10.
\displaystyle a=5.
\displaystyle S_m=\frac{m}{2}[2a+(m-1)d].
\displaystyle =\frac{m}{2}[2\times5+(m-1)\times2].
\displaystyle =\frac{m}{2}[10+2m-2].
\displaystyle =\frac{m}{2}(2m+8).
\displaystyle =m(m+4)=m^2+4m.

\displaystyle \textbf{Question 33: }\text{Find the sum of first 24 terms of an AP, whose }n\text{th term is given by } \\ a_n=3+2n.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a_n=3+2n.
\displaystyle \text{Sum of first 24 terms: }S_{24}=\frac{24}{2}(a_1+a_{24}).
\displaystyle a_1=3+2\times1=5\text{ and }a_{24}=3+2\times24=51.
\displaystyle S_{24}=12(5+51)=12\times56=672.

\displaystyle \textbf{Question 34: }\text{If the sum of first }n\text{ terms of an AP is given by }S_n=6n+7n^2,
\displaystyle \text{then find the }n\text{th term of an AP. Also, find 10th term of an AP.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }S_n=6n+7n^2.\quad (i)
\displaystyle S_{n-1}=6(n-1)+7(n-1)^2.
\displaystyle =6n-6+7(n^2-2n+1).
\displaystyle =6n-6+7n^2-14n+7.
\displaystyle =7n^2-8n+1.\quad (ii)
\displaystyle a_n=S_n-S_{n-1}.
\displaystyle =6n+7n^2-(7n^2-8n+1).
\displaystyle =6n+7n^2-7n^2+8n-1.
\displaystyle =14n-1.
\displaystyle a_{10}=14\times10-1=140-1=139.

\displaystyle \textbf{Question 35: }\text{Examine that the list of numbers obtained from the following situation,} \\ \text{will be in the form of GP.}  \text{An insect starts from a point and travels in a straight path } \\ 1\text{ mm} \text{ in first second and half of the distance} \text{covered in previous second in the} \\ \text{succeeding second.}
\displaystyle \text{Answer:}

\displaystyle \text{Given, distance moved by the insect in 1st second }=1\text{ mm}.
\displaystyle \text{Distance moved by the insect in 2nd second }=\frac{1}{2}\text{ mm}.
\displaystyle \text{and so on distance moved by the insect in 3rd second }=\frac{1}{4}\text{ mm}.
\displaystyle \text{Here, }\frac{a_2}{a_1}=\frac{\frac{1}{2}}{1}=\frac{1}{2}.
\displaystyle \frac{a_3}{a_2}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}\text{ and so on.}
\displaystyle \text{Thus, }\frac{a_{k+1}}{a_k}\text{ is same for all }k\in N.
\displaystyle \text{So, the above list of numbers forms a GP.}

\displaystyle \textbf{Question 36: }\text{If }a,b\text{ and }c\text{ are in GP, then find the value of }\frac{a-b}{b-c}.
\displaystyle \text{Answer:}

\displaystyle \text{Given that, }a,b\text{ and }c\text{ are in GP.}
\displaystyle \text{Then, }\frac{b}{a}=\frac{c}{b}=r\text{ (constant)}.\quad (i)
\displaystyle b=ar\text{ and }c=br.
\displaystyle b=ar\text{ and }c=(ar)r=ar^2.
\displaystyle \frac{a-b}{b-c}=\frac{a-ar}{ar-ar^2}=\frac{a(1-r)}{ar(1-r)}=\frac{1}{r}.
\displaystyle \frac{a-b}{b-c}=\frac{1}{r}=\frac{a}{b}=\frac{b}{c}.\quad [\text{Using (i)}]

\displaystyle \textbf{Question 37: }\text{Find the }n\text{th term and the 12th term of the sequence }-6,18,-54,\ldots
\displaystyle \text{Answer:}

\displaystyle \text{Given, sequence is }-6,18,-54,\ldots
\displaystyle \frac{a_2}{a_1}=\frac{18}{-6}=-3\text{ and }\frac{a_3}{a_2}=\frac{-54}{18}=-3.
\displaystyle \text{and so on i.e. }\frac{a_{k+1}}{a_k}\text{ is same for all }k\in N.
\displaystyle \text{Clearly, the ratio of each term by its preceding term is same.}
\displaystyle \text{So, the given sequence forms a GP with first term }a=-6
\displaystyle \text{and common ratio }r=-3.
\displaystyle a_n=ar^{n-1}=-6(-3)^{n-1}=(-1)^n\cdot6\cdot3^{n-1}.
\displaystyle a_{12}=(-1)^{12}\cdot6\cdot3^{11}=6\cdot3^{11}=2\cdot3^{12}.
\displaystyle \text{Hence, the }n\text{th term of given GP is }(-1)^n\cdot6\cdot3^{n-1}
\displaystyle \text{and 12th term is }2\cdot3^{12}.

\displaystyle \textbf{Question 38: }\text{Which term of the GP }5,10,20,40,\ldots\text{ is }5120?
\displaystyle \text{Answer:}

\displaystyle \text{Given, GP is }5,10,20,40,\ldots
\displaystyle a=5\text{ and }r=\frac{10}{5}=2.
\displaystyle \text{Let }n\text{th term of given GP }=5120\text{ i.e. }a_n=5120.
\displaystyle a_n=ar^{n-1}=5120.
\displaystyle 5(2)^{n-1}=5120.
\displaystyle 2^{n-1}=\frac{5120}{5}=1024.
\displaystyle 2^{n-1}=1024\Rightarrow 2^{n-1}=2^{10}.
\displaystyle n-1=10\Rightarrow n=11.
\displaystyle \text{Hence, 11th term of given GP is }5120.

\displaystyle \textbf{Question 39: }\text{If the 4th and 9th terms of a GP are }54\text{ and }13122\text{ respectively, then find the GP.}
\displaystyle \text{Answer:}

\displaystyle \text{Let }a\text{ be the first term and }r\text{ be the common ratio of GP.}
\displaystyle a_4=54\Rightarrow ar^{4-1}=ar^3=54.\quad (i)
\displaystyle a_9=13122\Rightarrow ar^{9-1}=ar^8=13122.\quad (ii)
\displaystyle \text{Divide (ii) by (i): }\frac{ar^8}{ar^3}=\frac{13122}{54}.
\displaystyle r^5=243=3^5.
\displaystyle r=3.
\displaystyle \text{Put }r=3\text{ in (i): }a(3)^3=54\Rightarrow 27a=54.
\displaystyle a=\frac{54}{27}=2.
\displaystyle \text{Required GP is }a,ar,ar^2,ar^3,\ldots\text{ i.e. }2,6,18,54,\ldots

\displaystyle \textbf{Question 40: }\text{If the }p\text{th and }q\text{th terms of a GP are }q\text{ and }p\text{ respectively, then} \\ \text{show that its }(p+q)\text{th term is }  \left(\frac{q^p}{p^q}\right)^{\frac{1}{p-q}}.
\displaystyle \text{Answer:}

\displaystyle \text{Let the first term and common ratio of GP be }a\text{ and }r\text{ respectively.}
\displaystyle \text{According to the question, }a_q=ar^{q-1}=q.\quad (i)
\displaystyle \text{and }a_p=ar^{p-1}=p.\quad (ii)
\displaystyle \text{Divide (i) by (ii): }\frac{ar^{q-1}}{ar^{p-1}}=\frac{q}{p}.
\displaystyle r^{q-p}=\frac{q}{p}\Rightarrow r=\left(\frac{q}{p}\right)^{\frac{1}{q-p}}.
\displaystyle \text{Substitute }r\text{ in (i): }a\left(\frac{q}{p}\right)^{\frac{q-1}{q-p}}=q.
\displaystyle a=q\left(\frac{q}{p}\right)^{\frac{1-q}{q-p}} =q\left(\frac{p}{q}\right)^{\frac{q-1}{q-p}}.
\displaystyle \text{Now, }(p+q)\text{th term }a_{p+q}=ar^{p+q-1}.
\displaystyle =q\left(\frac{p}{q}\right)^{\frac{q-1}{q-p}} \left(\frac{q}{p}\right)^{\frac{p+q-1}{q-p}}.
\displaystyle =q\left(\frac{q}{p}\right)^{\frac{p}{q-p}} =\left(\frac{q^p}{p^q}\right)^{\frac{1}{p-q}}.
\displaystyle =q^{\frac{1-p}{p-q}+\frac{p+q-1}{p-q}}
\displaystyle =q^{\frac{p-q-p+1+p+q-1}{p-q}}
\displaystyle =q^{\frac{p}{p-q}}
\displaystyle \frac{q^{\frac{p}{p-q}}}{p^{\frac{p}{p-q}}}
\displaystyle =\left(\frac{q^p}{p^p}\right)^{\frac{1}{p-q}}
\displaystyle =\left(\frac{q^p}{p^q}\right)^{\frac{1}{p-q}}
\displaystyle \text{Hence proved.}

\displaystyle \textbf{Question 41: }\text{Find four numbers forming a GP in which the third term is greater than the} \\ \text{first term by }9   \text{and the second term is greater than 4th by }18.
\displaystyle \text{Answer:}

\displaystyle \text{Let the GP be }a,ar,ar^2,ar^3.
\displaystyle \text{Given, third term = first term }+9.
\displaystyle ar^2=a+9\Rightarrow ar^2-a=9.\quad (i)
\displaystyle \text{Also, second term = fourth term }+18.
\displaystyle ar=ar^3+18\Rightarrow ar-ar^3=18.\quad (ii)
\displaystyle \text{Divide (i) by (ii): }\frac{ar^2-a}{ar-ar^3}=\frac{9}{18}.
\displaystyle \frac{a(r^2-1)}{ar(1-r^2)}=\frac{1}{2}.
\displaystyle \frac{r^2-1}{r(1-r^2)}=\frac{1}{2}.
\displaystyle \frac{-1(1-r^2)}{r(1-r^2)}=\frac{1}{2}.
\displaystyle -\frac{1}{r}=\frac{1}{2}\Rightarrow r=-2.
\displaystyle \text{Substitute }r=-2\text{ in (i): }a(-2)^2-a=9.
\displaystyle 4a-a=9\Rightarrow 3a=9\Rightarrow a=3.
\displaystyle \text{GP is }3,3(-2),3(-2)^2,3(-2)^3.
\displaystyle \text{Hence, required four numbers are }3,-6,12,-24.

\displaystyle \textbf{Question 42: }\text{Let }S\text{ be the sum, }P\text{ be the product and }R\text{ be the sum of} \\ \text{the reciprocals of }3\text{ terms of a GP.}   \text{Then, find }P^2R^3:S^3.
\displaystyle \text{Answer:}

\displaystyle \text{Let GP be }\frac{a}{r},a,ar.
\displaystyle S=\frac{a}{r}+a+ar=\frac{a(r^2+r+1)}{r}.
\displaystyle P=\left(\frac{a}{r}\right)(a)(ar)=a^3.
\displaystyle R=\frac{r}{a}+\frac{1}{a}+\frac{1}{ar}=\frac{1}{a}\left(\frac{r^2+r+1}{r}\right).
\displaystyle \text{Now, }\frac{P^2R^3}{S^3}=\frac{a^6\cdot\frac{1}{a^3}\left(\frac{r^2+r+1}{r}\right)^3}{\left(\frac{a(r^2+r+1)}{r}\right)^3}.
\displaystyle =\frac{a^3\left(\frac{r^2+r+1}{r}\right)^3}{a^3\left(\frac{r^2+r+1}{r}\right)^3}=1.
\displaystyle \text{Therefore, required ratio is }1:1.

\displaystyle \textbf{Question 43: }\text{The }(m+n)\text{th and }(m-n)\text{th terms of a GP are }p\text{ and }q\text{ respectively.}
\displaystyle \text{Show that the }m\text{th and }n\text{th terms are }\sqrt{pq}\text{ and }p\left(\frac{q}{p}\right)^{\frac{m}{2n}},\text{ respectively.}
\displaystyle \text{Answer:}

\displaystyle \text{Let }a\text{ be the first term and }r\text{ be the common ratio.}
\displaystyle a_{m+n}=p\text{ and }a_{m-n}=q.
\displaystyle ar^{m+n-1}=p\text{ and }ar^{m-n-1}=q.
\displaystyle \frac{ar^{m+n-1}}{ar^{m-n-1}}=\frac{p}{q}.
\displaystyle r^{2n}=\frac{p}{q}.
\displaystyle r=\left(\frac{p}{q}\right)^{\frac{1}{2n}}.
\displaystyle \frac{1}{r}=\left(\frac{q}{p}\right)^{\frac{1}{2n}}.
\displaystyle \text{Now, }a_m=ar^{m-1}.
\displaystyle a_m=ar^{m+n-1}\left(\frac{1}{r}\right)^n.
\displaystyle a_m=a_{m+n}\left(\frac{1}{r}\right)^n.
\displaystyle a_m=p\left(\frac{q}{p}\right)^{\frac{n}{2n}}.
\displaystyle a_m=p\left(\frac{q}{p}\right)^{\frac{1}{2}}=\sqrt{pq}.
\displaystyle \text{Also, }a_n=ar^{n-1}.
\displaystyle a_n=ar^{m+n-1}\left(\frac{1}{r}\right)^m.
\displaystyle a_n=a_{m+n}\left(\frac{1}{r}\right)^m.
\displaystyle a_n=p\left(\frac{q}{p}\right)^{\frac{m}{2n}}.
\displaystyle \text{Hence proved.}

\displaystyle \textbf{Question 44: }\text{Find the sum of the series }2+6+18+54+\ldots+4374.
\displaystyle \text{Answer:}
\displaystyle \text{Given series is }2+6+18+54+\ldots+4374.
\displaystyle \text{Here }a=2,\ r=\frac{6}{2}=3>1.
\displaystyle \text{Required sum }=\frac{l r-a}{r-1}=\frac{4374\times3-2}{3-1}.
\displaystyle =\frac{13122-2}{2}=\frac{13120}{2}=6560.
\displaystyle \text{Hence, sum of given series is }6560.

\displaystyle \textbf{Question 45: }\text{Find the sum to }n\text{ terms of the sequence given by }a_n=2^n+3n,\ n\in N.
\displaystyle \text{Answer:}
\displaystyle \text{Let }S_n\text{ denotes the sum to }n\text{ terms of the given sequence. Then,}
\displaystyle S_n=a_1+a_2+a_3+\ldots+a_n.
\displaystyle S_n=(2^1+3\times1)+(2^2+3\times2)+(2^3+3\times3)+\ldots+(2^n+3\times n).
\displaystyle S_n=(2^1+2^2+2^3+\ldots+2^n)+(3\times1+3\times2+3\times3+\ldots+3\times n).
\displaystyle S_n=(2^1+2^2+2^3+\ldots+2^n)+3(1+2+3+\ldots+n).
\displaystyle =2\left(\frac{2^n-1}{2-1}\right)+3\left[\frac{n}{2}(1+n)\right].
\displaystyle =2(2^n-1)+\frac{3n}{2}(n+1).

\displaystyle \textbf{Question 46: }\text{Find the sum of the following sequences.}
\displaystyle \text{(i) }\frac{1}{2},\frac{3}{2},\frac{9}{2},\ldots\text{ upto 10 terms.}
\displaystyle \text{(ii) }2,-\frac{1}{2},\frac{1}{8},\ldots\text{ upto 12 terms.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given sequence is }\frac{1}{2},\frac{3}{2},\frac{9}{2},\ldots
\displaystyle \text{Here, }\frac{a_2}{a_1}=\frac{\frac{3}{2}}{\frac{1}{2}}=3,\ \frac{a_3}{a_2}=\frac{\frac{9}{2}}{\frac{3}{2}}=3\text{ and so on.}
\displaystyle \text{i.e. }\frac{a_{k+1}}{a_k}\text{ is same for all }k\in N.
\displaystyle \text{Clearly, the ratio of any term by its preceding is same.}
\displaystyle \text{So, the above sequence forms a GP, with first term }a=\frac{1}{2}\text{ and common ratio }r=3>1.
\displaystyle S_{10}=\frac{a(r^{10}-1)}{r-1}.
\displaystyle =\frac{\frac{1}{2}(3^{10}-1)}{3-1}.
\displaystyle =\frac{1}{2}\cdot\frac{3^{10}-1}{2}=\frac{1}{4}(3^{10}-1).
\displaystyle =\frac{1}{4}(59049-1)=14762.
\displaystyle \text{(ii) Given, sequence is }2,-\frac{1}{2},\frac{1}{8},\ldots
\displaystyle \text{Here, }\frac{a_2}{a_1}=\frac{-\frac{1}{2}}{2}=-\frac{1}{4},\ \frac{a_3}{a_2}=\frac{\frac{1}{8}}{-\frac{1}{2}}=-\frac{1}{4}\text{ and so on.}
\displaystyle \text{i.e. }\frac{a_{k+1}}{a_k}\text{ is same for all }k\in N.
\displaystyle \text{Clearly, the ratio of any term by its preceding term is same.}
\displaystyle \text{So, the above sequence forms a GP with first term }a=2\text{ and common ratio }r=-\frac{1}{4}<1.
\displaystyle S_{12}=\frac{a(1-r^{12})}{1-r},\ r<1.
\displaystyle =\frac{2\left[1-\left(-\frac{1}{4}\right)^{12}\right]}{1-\left(-\frac{1}{4}\right)}.
\displaystyle =\frac{2\left[1-\frac{1}{4^{12}}\right]}{1+\frac{1}{4}}.
\displaystyle =\frac{2\left[1-\frac{1}{4^{12}}\right]}{\frac{5}{4}}=\frac{8}{5}\left[1-\frac{1}{4^{12}}\right].

\displaystyle \textbf{Question 47: }\text{Find the sum of series }\frac{3}{5}+\frac{4}{5^2}+\frac{3}{5^3}+\frac{4}{5^4}+\ldots\text{ to }2n\text{ terms.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }S=\frac{3}{5}+\frac{4}{5^2}+\frac{3}{5^3}+\frac{4}{5^4}+\ldots\text{ to }2n\text{ terms.}
\displaystyle S=\left[\frac{3}{5}+\frac{3}{5^3}+\ldots\text{ to }n\text{ terms}\right]+\left[\frac{4}{5^2}+\frac{4}{5^4}+\ldots\text{ to }n\text{ terms}\right].
\displaystyle =\frac{3}{5}\left[\frac{1-\left(\frac{1}{5^2}\right)^n}{1-\frac{1}{5^2}}\right]+\frac{4}{5^2}\left[\frac{1-\left(\frac{1}{5^2}\right)^n}{1-\frac{1}{5^2}}\right].
\displaystyle =\frac{3}{5}\left[\frac{1-\frac{1}{5^{2n}}}{1-\frac{1}{25}}\right]+\frac{4}{25}\left[\frac{1-\frac{1}{5^{2n}}}{1-\frac{1}{25}}\right].
\displaystyle =\frac{3}{5}\left[\frac{1-\frac{1}{5^{2n}}}{\frac{24}{25}}\right]+\frac{4}{25}\left[\frac{1-\frac{1}{5^{2n}}}{\frac{24}{25}}\right].
\displaystyle =\frac{3}{5}\cdot\frac{25}{24}\left(1-\frac{1}{5^{2n}}\right)+\frac{4}{25}\cdot\frac{25}{24}\left(1-\frac{1}{5^{2n}}\right).
\displaystyle =\frac{5}{8}\left(1-\frac{1}{5^{2n}}\right)+\frac{1}{6}\left(1-\frac{1}{5^{2n}}\right).
\displaystyle =\left(1-\frac{1}{5^{2n}}\right)\left(\frac{5}{8}+\frac{1}{6}\right)=\left(1-\frac{1}{5^{2n}}\right)\frac{19}{24}.
\displaystyle =\frac{19}{24}\left(1-\frac{1}{5^{2n}}\right).

\displaystyle \textbf{Question 48: }\text{Find the sum of the products of the corresponding terms of the sequences}
\displaystyle 2,4,8,16,32\text{ and }128,32,8,2,\frac{1}{2}.
\displaystyle \text{Answer:}
\displaystyle \text{Given sequences }2,4,8,16,32\text{ and }128,32,8,2,\frac{1}{2}.
\displaystyle \text{Multiplying corresponding terms gives }256,128,64,32,16.
\displaystyle \text{Here }a=256,\ r=\frac{128}{256}=\frac{1}{2},\ |r|<1.
\displaystyle S_5=\frac{a(1-r^5)}{1-r}=\frac{256\left(1-\left(\frac{1}{2}\right)^5\right)}{1-\frac{1}{2}}.
\displaystyle =\frac{256\left(1-\frac{1}{32}\right)}{\frac{1}{2}}=512\left(1-\frac{1}{32}\right).
\displaystyle =512\cdot\frac{31}{32}=16\times31=496.

\displaystyle \textbf{Question 49: }\text{A GP consists of an even number of terms. If the sum of all the terms is }5\text{ times}
\displaystyle \text{the sum of terms occupying odd places, then find its common ratio.}
\displaystyle \text{Answer:}
\displaystyle \text{Let GP be }a,ar,ar^2,\ldots,ar^{2n-1}.
\displaystyle \text{Sum of all terms }=\frac{a(r^{2n}-1)}{r-1}.
\displaystyle \text{Sum of odd place terms }=\frac{a(r^{2n}-1)}{r^2-1}.
\displaystyle \text{Given }\frac{a(r^{2n}-1)}{r-1}=5\cdot\frac{a(r^{2n}-1)}{r^2-1}.
\displaystyle \frac{1}{r-1}=\frac{5}{(r-1)(r+1)}.
\displaystyle r+1=5\Rightarrow r=4.

\displaystyle \textbf{Question 50: }\text{If }S_1,S_2\text{ and }S_3\text{ are respectively the sum of }n,2n\text{ and }3n\text{ terms of a GP,}
\displaystyle \text{then prove that }S_1(S_3-S_2)=(S_2-S_1)^2.
\displaystyle \text{Answer:}
\displaystyle \text{Let first term }a\text{ and ratio }r.
\displaystyle S_1=\frac{a(1-r^n)}{1-r},\ S_2=\frac{a(1-r^{2n})}{1-r},\ S_3=\frac{a(1-r^{3n})}{1-r}.
\displaystyle S_1(S_3-S_2)=\frac{a(1-r^n)}{1-r}\cdot\frac{a(r^{2n}-r^{3n})}{1-r}.
\displaystyle =\frac{a^2r^{2n}(1-r^n)^2}{(1-r)^2}.\quad (i)
\displaystyle S_2-S_1=\frac{a(r^n-r^{2n})}{1-r}=\frac{ar^n(1-r^n)}{1-r}.
\displaystyle (S_2-S_1)^2=\frac{a^2r^{2n}(1-r^n)^2}{(1-r)^2}.\quad (ii)
\displaystyle \text{From (i) and (ii), }S_1(S_3-S_2)=(S_2-S_1)^2.

\displaystyle \textbf{Question 51: }\text{How many terms of GP }3,\frac{3}{2},\frac{3}{4},\ldots\text{ are needed to give the sum }\frac{3069}{512}\ ?
\displaystyle \text{Answer:}
\displaystyle \text{Given GP is }3,\frac{3}{2},\frac{3}{4},\ldots
\displaystyle a=3,\ r=\frac{1}{2},\ |r|<1.
\displaystyle S_n=\frac{a(1-r^n)}{1-r}=\frac{3\left(1-\left(\frac{1}{2}\right)^n\right)}{1-\frac{1}{2}}.
\displaystyle =6\left(1-\frac{1}{2^n}\right).
\displaystyle 6\left(1-\frac{1}{2^n}\right)=\frac{3069}{512}.
\displaystyle 1-\frac{1}{2^n}=\frac{3069}{3072}=\frac{1023}{1024}.
\displaystyle \frac{1}{2^n}=\frac{1}{1024}\Rightarrow 2^n=1024\Rightarrow n=10.
\displaystyle 6\left(1-\frac{1}{2^n}\right)=\frac{3069}{512}\Rightarrow 1-\frac{1}{2^n}=\frac{3069}{3072}.
\displaystyle \frac{1}{2^n}=1-\frac{3069}{3072}=\frac{3072-3069}{3072}=\frac{3}{3072}=\frac{1}{1024}.
\displaystyle 2^n=1024=2^{10}\Rightarrow n=10.
\displaystyle \text{Hence, }10\text{ terms are needed.}

\displaystyle \textbf{Question 52: }\text{At the end of each year, the value of a certain machine has depreciated by }20\%\text{ of its value}
\displaystyle \text{at the beginning of that year. If its initial value was }\text{Rs }1250,\text{ find the value at the end of }5\text{ yr.}
\displaystyle \text{Answer:}
\displaystyle \text{Given yearly depreciation }=20\%.
\displaystyle \text{Value each year }=80\%\text{ of previous year }=\frac{80}{100}=0.8.
\displaystyle a=1250,\ r=0.8.
\displaystyle a_6=a r^5=1250(0.8)^5.
\displaystyle =1250(0.32768)=409.6.
\displaystyle \text{Hence, value at end of }5\text{ years is }409.6.

\displaystyle \textbf{Question 53: }\text{The sum of the first three terms of a GP is }16\text{ and the sum of next three terms is }128.
\displaystyle \text{Determine the first term, the common ratio and the sum to }n\text{ terms of the GP.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the GP be }a,ar,ar^2,ar^3,\ldots
\displaystyle \text{According to the given condition,}
\displaystyle \text{Sum of first three terms }=a+ar+ar^2=16.\quad (i)
\displaystyle \text{and sum of next three terms }=ar^3+ar^4+ar^5=128.\quad (ii)
\displaystyle \text{On dividing Eq. (i) by Eq. (ii), we get}
\displaystyle \frac{a+ar+ar^2}{ar^3+ar^4+ar^5}=\frac{16}{128}.
\displaystyle \frac{a(1+r+r^2)}{ar^3(1+r+r^2)}=\frac{1}{8}.
\displaystyle \frac{1}{r^3}=\frac{1}{8}.
\displaystyle \left(\frac{1}{r}\right)^3=\left(\frac{1}{2}\right)^3.
\displaystyle \text{On comparing the base of power 3 from both sides, we get}
\displaystyle \frac{1}{r}=\frac{1}{2}\Rightarrow r=2.
\displaystyle \text{On putting }r=2\text{ in Eq. (i), we get}
\displaystyle a+2a+4a=16.
\displaystyle 7a=16\Rightarrow a=\frac{16}{7}.
\displaystyle \text{Now, sum to }n\text{ terms,}
\displaystyle S_n=\frac{a(r^n-1)}{r-1},\ r=2>1.
\displaystyle =\frac{\frac{16}{7}(2^n-1)}{2-1}.
\displaystyle =\frac{16}{7}(2^n-1).
\displaystyle \text{Hence, }a=\frac{16}{7},\ r=2\text{ and }S_n=\frac{16}{7}(2^n-1).

\displaystyle \textbf{Question 54: }\text{Find the least value of }n\text{ for which the sum }1+3+3^2+\ldots\text{ to }n\text{ terms is greater than }7000.
\displaystyle \text{Answer:}
\displaystyle \text{We have }S_n=1+3+3^2+\ldots\text{ to }n\text{ terms.}
\displaystyle a=1,\ r=3>1.
\displaystyle S_n=\frac{a(r^n-1)}{r-1}=\frac{3^n-1}{2}.
\displaystyle \frac{3^n-1}{2}>7000\Rightarrow 3^n-1>14000.
\displaystyle 3^n>14001.
\displaystyle n\log3>\log14001.
\displaystyle n>\frac{\log14001}{\log3}=\frac{4.1461}{0.4771}\approx8.69.
\displaystyle \text{Hence, least value of }n\text{ is }9.

\displaystyle \textbf{Question 55: }\text{The lengths of three unequal edges of a rectangular solid block are in GP.}
\displaystyle \text{If the volume of the block is }216\ \text{cm}^3\text{ and the total surface area is }252\ \text{cm}^2,
\displaystyle \text{then find the length of its edges.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the length, breadth and height be }\frac{a}{r},a,ar.
\displaystyle \text{Volume }=\frac{a}{r}\cdot a\cdot ar=216.
\displaystyle a^3=216=6^3\Rightarrow a=6.
\displaystyle \text{Surface area }=2(lb+bh+hl)=252.
\displaystyle 2\left(\frac{a}{r}\cdot a+a\cdot ar+\frac{a}{r}\cdot ar\right)=252.
\displaystyle 2a^2\left(\frac{1}{r}+r+1\right)=252.
\displaystyle 2(36)\left(\frac{1}{r}+r+1\right)=252.
\displaystyle \frac{1}{r}+r+1=\frac{252}{72}=\frac{7}{2}.
\displaystyle \text{Multiply by }r:\ 1+r^2+r=\frac{7}{2}r.
\displaystyle 2+2r^2+2r=7r.
\displaystyle 2r^2-5r+2=0.
\displaystyle (2r-1)(r-2)=0.
\displaystyle r=\frac{1}{2}\text{ or }r=2.
\displaystyle \text{If }r=\frac{1}{2},\ \text{length }=\frac{6}{\frac{1}{2}}=12\text{ cm},\ \text{breadth }=6\text{ cm},\ \text{height }=6\cdot\frac{1}{2}=3\text{ cm}.
\displaystyle \text{If }r=2,\ \text{length }=\frac{6}{2}=3\text{ cm},\ \text{breadth }=6\text{ cm},\ \text{height }=6\cdot2=12\text{ cm}.
\displaystyle \text{Hence, edges are }(12,6,3)\text{ cm or }(3,6,12)\text{ cm}.

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