\displaystyle \textbf{Question 1. }\text{If in an infinite G.P., first term is equal to }10\text{ times the sum of all}
\displaystyle \text{successive terms, then its common ratio is}
\displaystyle \text{(a) }\frac{1}{10}\qquad \text{(b) }\frac{1}{11}\qquad \text{(c) }\frac{1}{9}\qquad \text{(d) }\frac{1}{20}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and common ratio be }r.
\displaystyle \text{Sum of all successive terms}=ar+ar^2+ar^3+\cdots=\frac{ar}{1-r}
\displaystyle a=10\cdot\frac{ar}{1-r}
\displaystyle 1-r=10r
\displaystyle 11r=1
\displaystyle r=\frac{1}{11}
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 2. }\text{If the first term of a G.P. }a_1,a_2,a_3,\ldots\text{ is unity such that }
\displaystyle 4a_2+5a_3\text{ is least, }\text{then the common ratio of G.P. is}
\displaystyle \text{(a) }-\frac{2}{5}\qquad \text{(b) }-\frac{3}{5}\qquad \text{(c) }\frac{2}{5}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle a_1=1,\quad a_2=r,\quad a_3=r^2
\displaystyle 4a_2+5a_3=4r+5r^2
\displaystyle =5\left(r+\frac{2}{5}\right)^2-\frac{4}{5}
\displaystyle \text{This is least when }r=-\frac{2}{5}.
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 3. }\text{If }a,b,c\text{ are in A.P. and }x,y,z\text{ are in G.P., then the value of }x^{b-c}y^{c-a}z^{a-b}\text{ is}
\displaystyle \text{(a) }0\qquad \text{(b) }1\qquad \text{(c) }xyz\qquad \text{(d) }x^ay^bz^c
\displaystyle \text{Answer:}
\displaystyle \text{Since }a,b,c\text{ are in A.P., let }a=b-d,\quad c=b+d.
\displaystyle \text{Since }x,y,z\text{ are in G.P., let }x=\frac{y}{r},\quad z=yr.
\displaystyle x^{b-c}y^{c-a}z^{a-b}=x^{-d}y^{2d}z^{-d}
\displaystyle =\left(\frac{y^2}{xz}\right)^d
\displaystyle \text{But }y^2=xz
\displaystyle \therefore x^{b-c}y^{c-a}z^{a-b}=1
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 4. }\text{The first three of four given numbers are in G.P. and their last three are in A.P.}
\displaystyle \text{with common difference }6.\text{ If first and fourth numbers are equal, then the first number is}
\displaystyle \text{(a) }2\qquad \text{(b) }4\qquad \text{(c) }6\qquad \text{(d) }8
\displaystyle \text{Answer:}
\displaystyle \text{Let the four numbers be }a,\ b,\ c,\ a.
\displaystyle \text{Last three are in A.P. with common difference }6.
\displaystyle c=b+6,\qquad a=c+6=b+12
\displaystyle \text{First three are in G.P., so }b^2=ac
\displaystyle b^2=(b+12)(b+6)
\displaystyle b^2=b^2+18b+72
\displaystyle b=-4
\displaystyle \therefore a=b+12=8
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 5. }\text{If }a,b,c\text{ are in G.P. and }a^{1/x}=b^{1/y}=c^{1/z},\text{ then }x,y,z\text{ are in}
\displaystyle \text{(a) AP}\qquad \text{(b) GP}\qquad \text{(c) HP}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let }a^{1/x}=b^{1/y}=c^{1/z}=k.
\displaystyle \therefore a=k^x,\quad b=k^y,\quad c=k^z
\displaystyle \text{Since }a,b,c\text{ are in G.P., }b^2=ac
\displaystyle (k^y)^2=k^x\cdot k^z
\displaystyle k^{2y}=k^{x+z}
\displaystyle 2y=x+z
\displaystyle \therefore x,y,z\text{ are in A.P.}
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 6. }\text{If }S\text{ be the sum, }P\text{ the product and }R\text{ be the sum of the reciprocals of }
\displaystyle n\text{ terms }\text{of a G.P., then }P^2\text{ is equal to}
\displaystyle \text{(a) }\frac{S}{R}\qquad \text{(b) }\frac{R}{S}\qquad \text{(c) }\left(\frac{R}{S}\right)^n\qquad \text{(d) }\left(\frac{S}{R}\right)^n
\displaystyle \text{Answer:}
\displaystyle \text{Let the }n\text{ terms of the G.P. be }a,\ ar,\ ar^2,\ldots,ar^{n-1}.
\displaystyle S=a(1+r+r^2+\cdots+r^{n-1})
\displaystyle R=\frac{1}{a}\left(1+\frac{1}{r}+\frac{1}{r^2}+\cdots+\frac{1}{r^{n-1}}\right)
\displaystyle R=\frac{S}{a^2r^{n-1}}
\displaystyle \therefore \frac{S}{R}=a^2r^{n-1}
\displaystyle P=a\cdot ar\cdot ar^2\cdots ar^{n-1}=a^nr^{\frac{n(n-1)}{2}}
\displaystyle \therefore P^2=a^{2n}r^{n(n-1)}=\left(a^2r^{n-1}\right)^n
\displaystyle \therefore P^2=\left(\frac{S}{R}\right)^n
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 7. }\text{The fractional value of }2.\overline{357}\text{ is}
\displaystyle \text{(a) }\frac{2355}{1001}\qquad \text{(b) }\frac{2379}{997}\qquad \text{(c) }\frac{2355}{999}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle x=2.\overline{357}
\displaystyle 1000x=2357.\overline{357}
\displaystyle 1000x-x=2357.\overline{357}-2.\overline{357}
\displaystyle 999x=2355
\displaystyle x=\frac{2355}{999}
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 8. }\text{If }p\text{th, }q\text{th and }r\text{th terms of an A.P. are in G.P., then the common ratio}
\displaystyle \text{of this G.P. is}
\displaystyle \text{(a) }\frac{p-q}{q-r}\qquad \text{(b) }\frac{q-r}{p-q}\qquad \text{(c) }pqr\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle T_p=a+(p-1)d,\quad T_q=a+(q-1)d,\quad T_r=a+(r-1)d
\displaystyle \text{Since these are in G.P.,}
\displaystyle \frac{T_q}{T_p}=\frac{T_r}{T_q}
\displaystyle T_q^2=T_pT_r
\displaystyle \text{On simplifying, common ratio}=\frac{T_q}{T_p}=\frac{q-r}{p-q}
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 9. }\text{The value of }9^{1/3}\cdot9^{1/9}\cdot9^{1/27}\cdots\text{ to }\infty,\text{ is}
\displaystyle \text{(a) }1\qquad \text{(b) }3\qquad \text{(c) }9\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle 9^{1/3}\cdot9^{1/9}\cdot9^{1/27}\cdots
\displaystyle =9^{\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\cdots\right)}
\displaystyle \frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\cdots=\frac{\frac{1}{3}}{1-\frac{1}{3}}=\frac{1}{2}
\displaystyle \therefore \text{Required value}=9^{1/2}=3
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 10. }\text{The sum of an infinite G.P. is }4\text{ and the sum of the cubes of its terms is }\frac{64}{7}. \\ \text{The common ratio of the original G.P. is}
\displaystyle \text{(a) }\frac{1}{2}\qquad \text{(b) }\frac{2}{3}\qquad \text{(c) }\frac{1}{3}\qquad \text{(d) }-\frac{1}{2}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and common ratio be }r
\displaystyle \frac{a}{1-r}=4
\displaystyle a=4(1-r)\qquad \text{(i)}
\displaystyle \frac{a^3}{1-r^3}=\frac{64}{7}
\displaystyle \frac{[4(1-r)]^3}{1-r^3}=\frac{64}{7}
\displaystyle \frac{64(1-r)^3}{(1-r)(1+r+r^2)}=\frac{64}{7}
\displaystyle \frac{(1-r)^2}{1+r+r^2}=\frac{1}{7}
\displaystyle 7(1-2r+r^2)=1+r+r^2
\displaystyle 6r^2-15r+6=0
\displaystyle 2r^2-5r+2=0
\displaystyle (2r-1)(r-2)=0
\displaystyle r=\frac{1}{2}\quad \text{or}\quad r=2
\displaystyle \text{Since }|r|<1\text{ for an infinite G.P., }r=\frac{1}{2}
\displaystyle \therefore\text{Correct option is (a).}
\\

\displaystyle \textbf{Question 11. }\text{If the sum of first two terms of an infinite G.P. is }1\text{ and every term is twice}
\displaystyle \text{the sum of all the successive terms, then its first term is}
\displaystyle \text{(a) }\frac{1}{3}\qquad \text{(b) }\frac{2}{3}\qquad \text{(c) }\frac{1}{4}\qquad \text{(d) }\frac{3}{4}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and common ratio be }r.
\displaystyle a+ar=1
\displaystyle \text{Every term is twice the sum of all successive terms.}
\displaystyle a=2(ar+ar^2+ar^3+\cdots)
\displaystyle a=2\cdot\frac{ar}{1-r}
\displaystyle 1-r=2r
\displaystyle r=\frac{1}{3}
\displaystyle a\left(1+\frac{1}{3}\right)=1
\displaystyle a=\frac{3}{4}
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Corrected Question 12. }\text{The }n^{\text{th}}\text{ term of a G.P. is }128\text{ and the sum of its } \\ n\text{ terms is }255.\text{ If its common ratio is }2,\text{ then its first term is}
\displaystyle \text{(a) }1\qquad \text{(b) }3\qquad \text{(c) }8\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a
\displaystyle \text{Given common ratio }r=2
\displaystyle T_n=ar^{n-1}=128
\displaystyle a\cdot2^{n-1}=128\qquad \text{(i)}
\displaystyle S_n=\frac{a(2^n-1)}{2-1}=255
\displaystyle a(2^n-1)=255\qquad \text{(ii)}
\displaystyle \text{From (i), }a\cdot2^n=256
\displaystyle \therefore 256-a=255
\displaystyle a=1
\displaystyle \therefore\text{First term}=1
\displaystyle \therefore\text{Correct option is (a).}
\\

\displaystyle \textbf{Question 13. }\text{If second term of a G.P. is }2\text{ and the sum of its infinite terms is } 8, \\ \text{then its first term is}
\displaystyle \text{(a) }\frac{1}{4}\qquad \text{(b) }\frac{1}{2}\qquad \text{(c) }2\qquad \text{(d) }4
\displaystyle \text{Answer:}
\displaystyle ar=2
\displaystyle \frac{a}{1-r}=8
\displaystyle a=8(1-r)
\displaystyle 8r(1-r)=2
\displaystyle 4r^2-4r+1=0
\displaystyle (2r-1)^2=0
\displaystyle r=\frac{1}{2}
\displaystyle a=4
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 14. }\text{If }a,b,c\text{ are in G.P. and }x,y\text{ are A.M.'s between }a,b\text{ and }b,c \\ \text{ respectively, then}
\displaystyle \text{(a) }\frac{1}{x}+\frac{1}{y}=2\qquad \text{(b) }\frac{1}{x}+\frac{1}{y}=\frac{1}{2}\qquad \text{(c) }\frac{1}{x}+\frac{1}{y}=\frac{2}{a}\qquad \text{(d) }\frac{1}{x}+\frac{1}{y}=\frac{2}{b}
\displaystyle \text{Answer:}
\displaystyle x=\frac{a+b}{2},\qquad y=\frac{b+c}{2}
\displaystyle \frac{1}{x}+\frac{1}{y}=\frac{2}{a+b}+\frac{2}{b+c}
\displaystyle =\frac{2(a+2b+c)}{(a+b)(b+c)}
\displaystyle \text{Since }a,b,c\text{ are in G.P., }b^2=ac
\displaystyle (a+b)(b+c)=ab+ac+b^2+bc=ab+2b^2+bc=b(a+2b+c)
\displaystyle \therefore \frac{1}{x}+\frac{1}{y}=\frac{2}{b}
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 15. }\text{If }A\text{ be one A.M. and }p,q\text{ be two G.M.'s between two numbers, then } \\ 2A\text{ is equal to}
\displaystyle \text{(a) }\frac{p^3+q^3}{pq}\qquad \text{(b) }\frac{p^3-q^3}{pq}\qquad \text{(c) }\frac{p^2+q^2}{2}\qquad \text{(d) }\frac{pq}{2}
\displaystyle \text{Answer:}
\displaystyle \text{Let the two numbers be }a\text{ and }b.
\displaystyle \text{Since }p,q\text{ are two G.M.'s, }a,p,q,b\text{ are in G.P.}
\displaystyle p=ar,\qquad q=ar^2,\qquad b=ar^3
\displaystyle a=\frac{p^2}{q},\qquad b=\frac{q^2}{p}
\displaystyle 2A=a+b
\displaystyle =\frac{p^2}{q}+\frac{q^2}{p}
\displaystyle =\frac{p^3+q^3}{pq}
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 16. }\text{If }p,q\text{ be two A.M.'s and }G\text{ be one G.M. between two numbers, then }G^2=
\displaystyle \text{(a) }(2p-q)(p-2q)\qquad \text{(b) }(2p-q)(2q-p)\qquad \text{(c) }(2p-q)(p+2q)\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let the two numbers be }a\text{ and }b.
\displaystyle \text{Since }p,q\text{ are two A.M.'s, }a,p,q,b\text{ are in A.P.}
\displaystyle p=a+d,\qquad q=a+2d
\displaystyle a=2p-q,\qquad b=a+3d=2q-p
\displaystyle G^2=ab
\displaystyle \therefore G^2=(2p-q)(2q-p)
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 17. }\text{If }x\text{ is positive, the sum to infinity of the series } \\ \frac{1}{1+x}-\frac{1-x}{(1+x)^2}+\frac{(1-x)^2}{(1+x)^3}
\displaystyle -\frac{(1-x)^3}{(1+x)^4}+\cdots\text{ is}
\displaystyle \text{(a) }\frac{1}{2}\qquad \text{(b) }\frac{3}{4}\qquad \text{(c) }1\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{This is an infinite G.P. with first term }\frac{1}{1+x}
\displaystyle \text{and common ratio }-\frac{1-x}{1+x}=\frac{x-1}{1+x}.
\displaystyle S_\infty=\frac{\frac{1}{1+x}}{1-\frac{x-1}{1+x}}
\displaystyle =\frac{\frac{1}{1+x}}{\frac{2}{1+x}}=\frac{1}{2}
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 18. }\text{If }(4^3)(4^6)(4^9)(4^{12})\cdots(4^{3x})=(0.0625)^{-54},\text{ the value of }x\text{ is}
\displaystyle \text{(a) }7\qquad \text{(b) }8\qquad \text{(c) }9\qquad \text{(d) }10
\displaystyle \text{Answer:}
\displaystyle (4^3)(4^6)(4^9)\cdots(4^{3x})=4^{3+6+9+\cdots+3x}
\displaystyle =4^{3(1+2+3+\cdots+x)}
\displaystyle =4^{\frac{3x(x+1)}{2}}
\displaystyle 0.0625=\frac{1}{16}=4^{-2}
\displaystyle (0.0625)^{-54}=(4^{-2})^{-54}=4^{108}
\displaystyle \frac{3x(x+1)}{2}=108
\displaystyle x(x+1)=72
\displaystyle x^2+x-72=0
\displaystyle (x-8)(x+9)=0
\displaystyle x=8
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 19. }\text{Given that }x>0,\text{ the sum }\sum_{n=1}^{\infty}\left(\frac{x}{x+1}\right)^{n-1}\text{ equals}
\displaystyle \text{(a) }x\qquad \text{(b) }x+1\qquad \text{(c) }\frac{x}{2x+1}\qquad \text{(d) }\frac{x+1}{2x+1}
\displaystyle \text{Answer:}
\displaystyle \sum_{n=1}^{\infty}\left(\frac{x}{x+1}\right)^{n-1}
\displaystyle =1+\frac{x}{x+1}+\left(\frac{x}{x+1}\right)^2+\cdots
\displaystyle \text{This is an infinite G.P. with common ratio }\frac{x}{x+1}.
\displaystyle \therefore S_\infty=\frac{1}{1-\frac{x}{x+1}}
\displaystyle =\frac{1}{\frac{1}{x+1}}=x+1
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 20. }\text{In a G.P. of even number of terms, the sum of all terms is five times the sum}
\displaystyle \text{of the odd terms. The common ratio of the G.P. is}
\displaystyle \text{(a) }-\frac{4}{5}\qquad \text{(b) }\frac{1}{5}\qquad \text{(c) }4\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let the G.P. be }a,ar,ar^2,\ldots\text{ with even number of terms.}
\displaystyle \text{Sum of all terms}=\text{sum of odd terms}+\text{sum of even terms}
\displaystyle \text{Sum of even terms}=r(\text{sum of odd terms})
\displaystyle \therefore \text{Sum of all terms}=(1+r)(\text{sum of odd terms})
\displaystyle 1+r=5
\displaystyle r=4
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 21. }\text{Let }x\text{ be the A.M. and }y,z\text{ be two G.M.s between two positive numbers. Then,}
\displaystyle \frac{y^3+z^3}{xyz}\text{ is equal to}
\displaystyle \text{(a) }1\qquad \text{(b) }2\qquad \text{(c) }\frac{1}{2}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let the two positive numbers be }a\text{ and }b.
\displaystyle x=\frac{a+b}{2}
\displaystyle \text{Since }y,z\text{ are two G.M.s, }a,y,z,b\text{ are in G.P.}
\displaystyle \text{Let }a=A,\quad y=Ar,\quad z=Ar^2,\quad b=Ar^3.
\displaystyle x=\frac{A+Ar^3}{2}=\frac{A(1+r^3)}{2}
\displaystyle \frac{y^3+z^3}{xyz}=\frac{A^3r^3+A^3r^6}{\frac{A(1+r^3)}{2}\cdot Ar\cdot Ar^2}
\displaystyle =\frac{A^3r^3(1+r^3)}{\frac{A^3r^3(1+r^3)}{2}}=2
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 22. }\text{The product }(32)(32)^{1/6}(32)^{1/36}\cdots\text{ to }\infty\text{ is equal to}
\displaystyle \text{(a) }64\qquad \text{(b) }16\qquad \text{(c) }32\qquad \text{(d) }0
\displaystyle \text{Answer:}
\displaystyle (32)(32)^{1/6}(32)^{1/36}\cdots
\displaystyle =32^{1+\frac{1}{6}+\frac{1}{36}+\cdots}
\displaystyle 1+\frac{1}{6}+\frac{1}{36}+\cdots=\frac{1}{1-\frac{1}{6}}=\frac{6}{5}
\displaystyle \therefore \text{Required product}=32^{6/5}=(2^5)^{6/5}=2^6=64
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 23. }\text{The two geometric means between the numbers }1\text{ and }64\text{ are}
\displaystyle \text{(a) }1\text{ and }64\qquad \text{(b) }4\text{ and }16\qquad \text{(c) }2\text{ and }16\qquad \text{(d) }8\text{ and }16
\displaystyle \text{Answer:}
\displaystyle \text{Let the numbers be }1,\ G_1,\ G_2,\ 64\text{ in G.P.}
\displaystyle 64=1\cdot r^3
\displaystyle r=4
\displaystyle \therefore G_1=4,\quad G_2=16
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 24. }\text{In a G.P. if the }(m+n)\text{th term is }p\text{ and }(m-n)\text{th term is } \\ q,\text{ then its }m\text{th term is}
\displaystyle \text{(a) }0\qquad \text{(b) }pq\qquad \text{(c) }\sqrt{pq}\qquad \text{(d) }\frac{1}{2}(p+q)
\displaystyle \text{Answer:}
\displaystyle T_{m+n}=p,\qquad T_{m-n}=q
\displaystyle \text{In a G.P., }T_m^2=T_{m+n}\cdot T_{m-n}
\displaystyle \therefore T_m^2=pq
\displaystyle \therefore T_m=\sqrt{pq}
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 25. }\text{Let }S\text{ be the sum, }P\text{ be the product and }R\text{ be the sum of the reciprocals of }
\displaystyle 3\text{ terms } \text{of a G.P. then }P^2R^3:S^3\text{ is equal to}
\displaystyle \text{(a) }1:1\qquad \text{(b) }(\text{Common ratio})^n:1\qquad \text{(c) }(\text{First term})^2(\text{Common ratio})^2\qquad \text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let the three terms of the G.P. be }\frac{a}{r},\ a,\ ar.
\displaystyle S=\frac{a}{r}+a+ar=a\left(\frac{1+r+r^2}{r}\right)
\displaystyle P=\frac{a}{r}\cdot a\cdot ar=a^3
\displaystyle R=\frac{r}{a}+\frac{1}{a}+\frac{1}{ar}=\frac{1+r+r^2}{ar}
\displaystyle \frac{S}{R}=a^2
\displaystyle \therefore P^2R^3:S^3=a^6R^3:S^3=a^6:\left(\frac{S}{R}\right)^3
\displaystyle =a^6:a^6=1:1
\displaystyle \therefore \text{Correct option is (a).}
\\

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