\displaystyle \text{MULTIPLE CHOICE QUESTIONS (MCQs)}

\displaystyle \textbf{Question 1: }\text{If }7^{th}\text{ and }13^{th}\text{ terms of an A.P. be }34\text{ and }64
\displaystyle \text{respectively, then its }18^{th}\text{ term is}
\displaystyle \text{(a) }87\qquad\text{(b) }88\qquad \text{(c) }89\qquad\text{(d) }90
\displaystyle \text{Answer:}
\displaystyle T_7=a+6d=34.
\displaystyle T_{13}=a+12d=64.
\displaystyle \text{Subtracting, }6d=30\Rightarrow d=5.
\displaystyle T_{18}=T_7+(18-7)d.
\displaystyle =34+11(5)=89.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If the sum of }p\text{ terms of an A.P. is }q\text{ and the sum of }q\text{ terms is }p,
\displaystyle \text{then the sum of }p+q\text{ terms will be}
\displaystyle \text{(a) }0\qquad\text{(b) }p-q\qquad \text{(c) }p+q\qquad\text{(d) }-(p+q)
\displaystyle \text{Answer:}
\displaystyle S_n=An^2+Bn.
\displaystyle S_p=Ap^2+Bp=q\Rightarrow Ap+B=\frac{q}{p}.
\displaystyle S_q=Aq^2+Bq=p\Rightarrow Aq+B=\frac{p}{q}.
\displaystyle \text{Subtracting, }A(p-q)=\frac{q}{p}-\frac{p}{q}.
\displaystyle A(p-q)=\frac{q^2-p^2}{pq} =-\frac{(p-q)(p+q)}{pq}.
\displaystyle \therefore A=-\frac{p+q}{pq}.
\displaystyle B=\frac{q}{p}-Ap =\frac{q}{p}+\frac{p+q}{q}.
\displaystyle S_{p+q}=A(p+q)^2+B(p+q).
\displaystyle =(p+q)\left[-\frac{(p+q)^2}{pq} +\frac{q}{p}+\frac{p+q}{q}\right].
\displaystyle =(p+q)\left[\frac{-(p+q)^2+q^2+p(p+q)}{pq}\right].
\displaystyle =(p+q)\left(\frac{-pq}{pq}\right)=-(p+q).
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If the sum of }n\text{ terms of an A.P. be }3n^2-n\text{ and its common}
\displaystyle \text{difference is }6,\text{ then its first term is}
\displaystyle \text{(a) }2\qquad\text{(b) }3\qquad \text{(c) }1\qquad\text{(d) }4
\displaystyle \text{Answer:}
\displaystyle S_n=3n^2-n.
\displaystyle \text{The first term is }a=S_1.
\displaystyle a=3(1)^2-1=2.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Sum of all two digit numbers which when divided by }4\text{ yield unity}
\displaystyle \text{as remainder is}
\displaystyle \text{(a) }1200\qquad\text{(b) }1210\qquad \text{(c) }1250\qquad\text{(d) none of these.}
\displaystyle \text{Answer:}
\displaystyle \text{The required two digit numbers are }13,17,21,\ldots,97.
\displaystyle \text{Here, }a=13,\quad d=4,\quad l=97.
\displaystyle 97=13+(n-1)4.
\displaystyle 4(n-1)=84\Rightarrow n=22.
\displaystyle S_{22}=\frac{22}{2}(13+97).
\displaystyle =11(110)=1210.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }n\text{ A.M.'s are introduced between }3\text{ and }17\text{ such that the ratio}
\displaystyle \text{of the last mean to the first mean is }3:1,\text{ then the value of }n\text{ is}
\displaystyle \text{(a) }6\qquad\text{(b) }8\qquad \text{(c) }4\qquad\text{(d) none of these.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the common difference be }d.
\displaystyle \text{Since }n\text{ A.M.'s are inserted between }3\text{ and }17,
\displaystyle d=\frac{17-3}{n+1}=\frac{14}{n+1}.
\displaystyle \text{First mean}=3+d,\qquad \text{last mean}=3+nd.
\displaystyle \frac{3+nd}{3+d}=3.
\displaystyle 3+nd=9+3d.
\displaystyle (n-3)d=6.
\displaystyle (n-3)\frac{14}{n+1}=6.
\displaystyle 14n-42=6n+6.
\displaystyle 8n=48\Rightarrow n=6.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }S_n\text{ denotes the sum of first }n\text{ terms of an A.P. }\langle a_n\rangle\text{ such that}
\displaystyle \frac{S_m}{S_n}=\frac{m^2}{n^2},\text{ then }\frac{a_m}{a_n}=
\displaystyle \text{(a) }\frac{2m+1}{2n+1}\qquad \text{(b) }\frac{2m-1}{2n-1}\qquad \text{(c) }\frac{m-1}{n-1}\qquad \text{(d) }\frac{m+1}{n+1}
\displaystyle \text{Answer:}
\displaystyle \frac{S_m}{S_n}=\frac{m^2}{n^2}\Rightarrow S_n=kn^2,\text{ for some constant }k.
\displaystyle a_n=S_n-S_{n-1}.
\displaystyle =kn^2-k(n-1)^2=k(2n-1).
\displaystyle \text{Similarly, }a_m=k(2m-1).
\displaystyle \therefore \frac{a_m}{a_n}=\frac{2m-1}{2n-1}.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The first and last terms of an A.P. are }1\text{ and }11.\text{ If the sum of its}
\displaystyle \text{terms is }36,\text{ then the number of terms will be}
\displaystyle \text{(a) }5\qquad\text{(b) }6\qquad \text{(c) }7\qquad\text{(d) }8
\displaystyle \text{Answer:}
\displaystyle \text{Here, }a=1,\quad l=11,\quad S_n=36.
\displaystyle S_n=\frac{n}{2}(a+l).
\displaystyle 36=\frac{n}{2}(1+11).
\displaystyle 36=6n\Rightarrow n=6.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If the sum of }n\text{ terms of an A.P. is }3n^2+5n,\text{ then which of its}
\displaystyle \text{terms is }164?
\displaystyle \text{(a) }26^{th}\qquad\text{(b) }27^{th}\qquad \text{(c) }28^{th}\qquad\text{(d) none of these.}
\displaystyle \text{Answer:}
\displaystyle S_n=3n^2+5n.
\displaystyle a_n=S_n-S_{n-1}.
\displaystyle =3n^2+5n-\{3(n-1)^2+5(n-1)\}.
\displaystyle =3n^2+5n-(3n^2-n-2).
\displaystyle =6n+2.
\displaystyle \text{For }a_n=164,
\displaystyle 6n+2=164.
\displaystyle 6n=162\Rightarrow n=27.
\displaystyle \therefore 164\text{ is the }27^{th}\text{ term.}
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If the sum of }n\text{ terms of an A.P. is }2n^2+5n,\text{ then its }n^{th}\text{ term is}
\displaystyle \text{(a) }4n-3\qquad\text{(b) }3n-4\qquad \text{(c) }4n+3\qquad\text{(d) }3n+4
\displaystyle \text{Answer:}
\displaystyle S_n=2n^2+5n.
\displaystyle a_n=S_n-S_{n-1}.
\displaystyle =2n^2+5n-\{2(n-1)^2+5(n-1)\}.
\displaystyle =2n^2+5n-(2n^2+n-3).
\displaystyle =4n+3.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }a_1,a_2,a_3,\ldots,a_n\text{ are in A.P. with common difference }d,\text{ then}
\displaystyle \text{the sum of the series }\sin d\,[\mathrm{cosec}\,a_1\mathrm{cosec}\,a_2+
\displaystyle \mathrm{cosec}\,a_1\mathrm{cosec}\,a_3+\cdots+\mathrm{cosec}\,a_{n-1}\mathrm{cosec}\,a_n]\text{ is}
\displaystyle \text{(a) }\sec a_1-\sec a_n\qquad \text{(b) }\mathrm{cosec}\,a_1-\mathrm{cosec}\,a_n
\displaystyle \text{(c) }\cot a_1-\cot a_n\qquad \text{(d) }\tan a_1-\tan a_n
\displaystyle \text{Answer:}
\displaystyle \text{Since }a_1,a_2,\ldots,a_n\text{ are in A.P., }a_{r+1}-a_r=d.
\displaystyle \sin d\,\mathrm{cosec}\,a_r\mathrm{cosec}\,a_{r+1} =\frac{\sin(a_{r+1}-a_r)}{\sin a_r\sin a_{r+1}}.
\displaystyle =\cot a_r-\cot a_{r+1}.
\displaystyle \therefore \text{Sum}=(\cot a_1-\cot a_2)+(\cot a_2-\cot a_3)+\cdots
\displaystyle +(\cot a_{n-1}-\cot a_n).
\displaystyle =\cot a_1-\cot a_n.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{In the arithmetic progression whose common difference is non-zero, the sum}
\displaystyle \text{of first }3n\text{ terms is equal to the sum of next }n\text{ terms. Then the ratio of the sum}
\displaystyle \text{of the first }2n\text{ terms to the next }2n\text{ terms is equal to}
\displaystyle \text{(a) }\frac15\qquad\text{(b) }\frac23\qquad \text{(c) }\frac34\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and common difference be }d.
\displaystyle S_{3n}=S_{4n}-S_{3n}\Rightarrow 2S_{3n}=S_{4n}.
\displaystyle 2\left[\frac{3n}{2}\{2a+(3n-1)d\}\right] =\frac{4n}{2}\{2a+(4n-1)d\}.
\displaystyle 3\{2a+(3n-1)d\}=2\{2a+(4n-1)d\}.
\displaystyle 6a+(9n-3)d=4a+(8n-2)d.
\displaystyle 2a+(n-1)d=0.
\displaystyle S_{2n}=\frac{2n}{2}\{2a+(2n-1)d\}=n^2d.
\displaystyle S_{4n}=2n\{2a+(4n-1)d\}=6n^2d.
\displaystyle \text{Sum of the next }2n\text{ terms}=S_{4n}-S_{2n}=5n^2d.
\displaystyle \therefore \frac{\text{sum of first }2n\text{ terms}} {\text{sum of next }2n\text{ terms}}=\frac{n^2d}{5n^2d}=\frac15.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }a_1,a_2,a_3,\ldots,a_n\text{ are in A.P. with common difference }d,\text{ then the sum of the series}
\displaystyle \sin d\,[\sec a_1\sec a_2+\sec a_2\sec a_3+\cdots+\sec a_{n-1}\sec a_n]\text{ is}
\displaystyle \text{(a) }\sec a_1-\sec a_n\qquad \text{(b) }\mathrm{cosec}\,a_1-\mathrm{cosec}\,a_n
\displaystyle \text{(c) }\cot a_1-\cot a_n\qquad \text{(d) }\tan a_n-\tan a_1
\displaystyle \text{Answer:}
\displaystyle \text{Since }a_1,a_2,\ldots,a_n\text{ are in A.P., }a_{r+1}-a_r=d.
\displaystyle \sin d\,\sec a_r\sec a_{r+1} =\frac{\sin(a_{r+1}-a_r)}{\cos a_r\cos a_{r+1}}.
\displaystyle =\tan a_{r+1}-\tan a_r.
\displaystyle \therefore \text{Sum}=(\tan a_2-\tan a_1)+(\tan a_3-\tan a_2)+\cdots
\displaystyle +(\tan a_n-\tan a_{n-1}).
\displaystyle =\tan a_n-\tan a_1.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If four numbers in A.P. are such that their sum is }50\text{ and the greatest}
\displaystyle \text{number is }4\text{ times the least, then the numbers are}
\displaystyle \text{(a) }5,10,15,20\qquad \text{(b) }4,10,16,22\qquad \text{(c) }3,7,11,15\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let the four numbers be }a-3d,\ a-d,\ a+d,\ a+3d.
\displaystyle (a-3d)+(a-d)+(a+d)+(a+3d)=50.
\displaystyle 4a=50\Rightarrow a=\frac{25}{2}.
\displaystyle \text{Also, }a+3d=4(a-3d).
\displaystyle 15d=3a\Rightarrow d=\frac{a}{5}=\frac52.
\displaystyle a-3d=5,\quad a-d=10,\quad a+d=15,\quad a+3d=20.
\displaystyle \therefore \text{The numbers are }5,10,15,20.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }n\text{ arithmetic means are inserted between }1\text{ and }31\text{ such that}
\displaystyle \text{the ratio of the first mean and }n^{th}\text{ mean is }3:29,\text{ then the value of }n\text{ is}
\displaystyle \text{(a) }10\qquad\text{(b) }12\qquad \text{(c) }13\qquad\text{(d) }14
\displaystyle \text{Answer:}
\displaystyle \text{Let the common difference be }d.
\displaystyle d=\frac{31-1}{n+1}=\frac{30}{n+1}.
\displaystyle \text{First mean}=1+d=1+\frac{30}{n+1}=\frac{n+31}{n+1}.
\displaystyle n^{th}\text{ mean}=1+nd=1+\frac{30n}{n+1}=\frac{31n+1}{n+1}.
\displaystyle \frac{n+31}{31n+1}=\frac{3}{29}.
\displaystyle 29(n+31)=3(31n+1).
\displaystyle 29n+899=93n+3.
\displaystyle 64n=896\Rightarrow n=14.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Let }S_n\text{ denote the sum of }n\text{ terms of an A.P. whose first term is }a.
\displaystyle \text{If the common difference }d\text{ is given by }d=S_n-kS_{n-1}+S_{n-2},\text{ then }k=
\displaystyle \text{(a) }1\qquad\text{(b) }2\qquad \text{(c) }3\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle a_n=S_n-S_{n-1}.
\displaystyle a_{n-1}=S_{n-1}-S_{n-2}.
\displaystyle \text{Since the common difference is }d=a_n-a_{n-1},
\displaystyle d=(S_n-S_{n-1})-(S_{n-1}-S_{n-2}).
\displaystyle d=S_n-2S_{n-1}+S_{n-2}.
\displaystyle \text{Comparing with }d=S_n-kS_{n-1}+S_{n-2},
\displaystyle k=2.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{The first and last term of an A.P. are }a\text{ and }l\text{ respectively. If }S\text{ is the sum}
\displaystyle \text{of all the terms of the A.P. and the common difference is given by }\frac{l^2-a^2}{k-(l+a)},\text{ then }k=
\displaystyle \text{(a) }S\qquad\text{(b) }2S\qquad \text{(c) }3S\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let the number of terms be }n.
\displaystyle l=a+(n-1)d.
\displaystyle \therefore d=\frac{l-a}{n-1}.
\displaystyle \text{Also, }S=\frac{n}{2}(a+l)\Rightarrow n=\frac{2S}{a+l}.
\displaystyle d=\frac{l-a}{\frac{2S}{a+l}-1}.
\displaystyle =\frac{(l-a)(l+a)}{2S-(l+a)} =\frac{l^2-a^2}{2S-(l+a)}.
\displaystyle \text{Comparing with }d=\frac{l^2-a^2}{k-(l+a)},
\displaystyle k=2S.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{If the sum of first }n\text{ even natural numbers is equal to }k\text{ times the sum}
\displaystyle \text{of first }n\text{ odd natural numbers, then }k=
\displaystyle \text{(a) }\frac{1}{n}\qquad \text{(b) }\frac{n-1}{n}\qquad \text{(c) }\frac{n+1}{2n}\qquad \text{(d) }\frac{n+1}{n}
\displaystyle \text{Answer:}
\displaystyle \text{Sum of first }n\text{ even natural numbers}=n(n+1).
\displaystyle \text{Sum of first }n\text{ odd natural numbers}=n^2.
\displaystyle n(n+1)=k n^2.
\displaystyle k=\frac{n(n+1)}{n^2}=\frac{n+1}{n}.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If the first, second and last term of an A.P. are }a,b\text{ and }2a
\displaystyle \text{respectively, then its sum is}
\displaystyle \text{(a) }\frac{ab}{2(b-a)}\qquad \text{(b) }\frac{ab}{b-a}\qquad \text{(c) }\frac{3ab}{2(b-a)}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Here, the common difference }d=b-a.
\displaystyle \text{Let the number of terms be }n.
\displaystyle 2a=a+(n-1)(b-a).
\displaystyle a=(n-1)(b-a).
\displaystyle n-1=\frac{a}{b-a}.
\displaystyle n=\frac{a+b-a}{b-a}=\frac{b}{b-a}.
\displaystyle S_n=\frac{n}{2}(\text{first term}+\text{last term}).
\displaystyle =\frac{1}{2}\frac{b}{b-a}(a+2a).
\displaystyle =\frac{3ab}{2(b-a)}.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If }S_1\text{ is the sum of an arithmetic progression of }n\text{ odd number of terms and }S_2
\displaystyle \text{the sum of the terms of the series in odd places, then }\frac{S_1}{S_2}=
\displaystyle \text{(a) }\frac{2n}{n+1}\qquad \text{(b) }\frac{n}{n+1}\qquad \text{(c) }\frac{n+1}{2n}\qquad \text{(d) }\frac{n+1}{n}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and common difference be }d.
\displaystyle S_1=\frac{n}{2}\{2a+(n-1)d\}.
\displaystyle \text{Since }n\text{ is odd, there are }\frac{n+1}{2}\text{ terms in odd places.}
\displaystyle \text{These terms are }a,\ a+2d,\ a+4d,\ldots,a+(n-1)d.
\displaystyle S_2=\frac{n+1}{4}\{2a+(n-1)d\}.
\displaystyle \therefore \frac{S_1}{S_2} =\frac{\frac{n}{2}\{2a+(n-1)d\}} {\frac{n+1}{4}\{2a+(n-1)d\}}.
\displaystyle =\frac{2n}{n+1}.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If in an A.P., }S_n=n^2p\text{ and }S_m=m^2p,\text{ where }S_r\text{ denotes}
\displaystyle \text{the sum of }r\text{ terms of the A.P., then }S_p\text{ is equal to}
\displaystyle \text{(a) }\frac{1}{2}p^3\qquad \text{(b) }mnp\qquad \text{(c) }p^3\qquad \text{(d) }(m+n)p^2
\displaystyle \text{Answer:}
\displaystyle \text{For an A.P., let }S_r=Ar^2+Br.
\displaystyle S_n=An^2+Bn=n^2p.
\displaystyle \therefore An+B=np.
\displaystyle S_m=Am^2+Bm=m^2p.
\displaystyle \therefore Am+B=mp.
\displaystyle \text{Subtracting, }A(n-m)=p(n-m)\Rightarrow A=p.
\displaystyle \therefore B=0.
\displaystyle \therefore S_r=pr^2.
\displaystyle S_p=p(p^2)=p^3.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If in an A.P., the }p^{th}\text{ term is }q\text{ and }(p+q)^{th}\text{ term is zero,}
\displaystyle \text{then the }q^{th}\text{ term is}
\displaystyle \text{(a) }-p\qquad \text{(b) }p\qquad \text{(c) }p+q\qquad \text{(d) }p-q
\displaystyle \text{Answer:}
\displaystyle T_p=q.
\displaystyle T_{p+q}=T_p+qd.
\displaystyle 0=q+qd.
\displaystyle q(1+d)=0\Rightarrow d=-1.
\displaystyle T_q=T_p+(q-p)d.
\displaystyle =q+(q-p)(-1).
\displaystyle =q-q+p=p.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{The }10^{th}\text{ common term between the A.P.s }3,7,11,15,\ldots\text{ and}
\displaystyle 1,6,11,16,\ldots\text{ is}
\displaystyle \text{(a) }191\qquad\text{(b) }193\qquad \text{(c) }211\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{The terms of the first A.P. are of the form }4r+3.
\displaystyle \text{The terms of the second A.P. are of the form }5s+1.
\displaystyle \text{The first common term is }11.
\displaystyle \text{The common difference between successive common terms is}
\displaystyle \mathrm{LCM}(4,5)=20.
\displaystyle \therefore \text{Common terms are }11,31,51,\ldots
\displaystyle T_{10}=11+(10-1)20.
\displaystyle =11+180=191.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{If in an A.P. }S_n=n^2q\text{ and }S_m=m^2q,\text{ where }S_r\text{ denotes}
\displaystyle \text{the sum of }r\text{ terms of the A.P., then }S_q\text{ equals}
\displaystyle \text{(a) }\frac{q^3}{2}\qquad \text{(b) }mnq\qquad \text{(c) }q^3\qquad \text{(d) }(m^2+n^2)q
\displaystyle \text{Answer:}
\displaystyle \text{For an A.P., let }S_r=Ar^2+Br.
\displaystyle S_n=An^2+Bn=n^2q.
\displaystyle \therefore An+B=nq.
\displaystyle S_m=Am^2+Bm=m^2q.
\displaystyle \therefore Am+B=mq.
\displaystyle \text{Subtracting, }A(n-m)=q(n-m)\Rightarrow A=q.
\displaystyle \therefore B=0.
\displaystyle \therefore S_r=qr^2.
\displaystyle S_q=q(q^2)=q^3.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{Let }S_n\text{ denote the sum of first }n\text{ terms of an A.P. If }S_{2n}=3S_n,
\displaystyle \text{then }S_{3n}:S_n\text{ is equal to}
\displaystyle \text{(a) }4\qquad\text{(b) }6\qquad \text{(c) }8\qquad\text{(d) }10
\displaystyle \text{Answer:}
\displaystyle \text{Let }S_r=Ar^2+Br.
\displaystyle S_{2n}=4An^2+2Bn,\qquad S_n=An^2+Bn.
\displaystyle \text{Given }S_{2n}=3S_n.
\displaystyle 4An^2+2Bn=3An^2+3Bn.
\displaystyle An^2=Bn\Rightarrow B=An.
\displaystyle S_n=An^2+Bn=2An^2.
\displaystyle S_{3n}=9An^2+3Bn=9An^2+3An^2=12An^2.
\displaystyle \therefore \frac{S_{3n}}{S_n} =\frac{12An^2}{2An^2}=6.
\displaystyle \therefore S_{3n}:S_n=6:1.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 25: }\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 the common ratio be }r.
\displaystyle \text{The sum of all successive terms after the first term is}
\displaystyle ar+ar^2+ar^3+\cdots=\frac{ar}{1-r}.
\displaystyle \text{Given, }a=10\left(\frac{ar}{1-r}\right).
\displaystyle 1=\frac{10r}{1-r}.
\displaystyle 1-r=10r.
\displaystyle 11r=1.
\displaystyle \therefore r=\frac{1}{11}.
\displaystyle \text{Verification: }\frac{ar}{1-r} =\frac{a/11}{1-1/11}=\frac{a}{10}.
\displaystyle \therefore a=10\left(\frac{a}{10}\right)=a.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 26: }\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, 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 \text{Since }a_1=1,\text{ let the common ratio be }r.
\displaystyle \therefore a_2=r,\qquad a_3=r^2.
\displaystyle 4a_2+5a_3=4r+5r^2.
\displaystyle =5\left(r^2+\frac{4}{5}r\right).
\displaystyle =5\left[\left(r+\frac{2}{5}\right)^2-\frac{4}{25}\right].
\displaystyle =5\left(r+\frac{2}{5}\right)^2-\frac{4}{5}.
\displaystyle \text{This expression is least when }\left(r+\frac{2}{5}\right)^2=0.
\displaystyle \therefore r=-\frac{2}{5}.
\displaystyle \text{Verification: since the coefficient of }r^2\text{ is }5>0,
\displaystyle \text{the value obtained gives the minimum of }4r+5r^2.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{If }a,b,c\text{ are in A.P. and }x,y,z\text{ are in G.P., then the value of}
\displaystyle 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 }b-a=c-b=d.
\displaystyle \therefore b-c=-d,\qquad c-a=2d,\qquad a-b=-d.
\displaystyle \text{Since }x,y,z\text{ are in G.P., }y^2=xz.
\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 =1^d=1.
\displaystyle \text{Verification: }y^2=xz\Rightarrow \frac{y^2}{xz}=1.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 28: }\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,D.
\displaystyle \text{Since }B,C,D\text{ are in A.P. with common difference }6,
\displaystyle C=B+6,\qquad D=B+12.
\displaystyle \text{Given that the first and fourth numbers are equal, }A=D=B+12.
\displaystyle \text{Since }A,B,C\text{ are in G.P., }B^2=AC.
\displaystyle B^2=(B+12)(B+6).
\displaystyle B^2=B^2+18B+72.
\displaystyle 18B+72=0\Rightarrow B=-4.
\displaystyle \therefore A=B+12=-4+12=8.
\displaystyle \text{Verification: the numbers are }8,-4,2,8.
\displaystyle \frac{-4}{8}=\frac{2}{-4}=-\frac12,\text{ so the first three are in G.P.}
\displaystyle 2-(-4)=6,\qquad 8-2=6,\text{ so the last three are in A.P.}
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 29: }\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) A.P.}\qquad \text{(b) G.P.}\qquad \text{(c) H.P.}\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,\qquad b=k^y,\qquad c=k^z.
\displaystyle \text{Since }a,b,c\text{ are in G.P., }b^2=ac.
\displaystyle (k^y)^2=k^xk^z.
\displaystyle k^{2y}=k^{x+z}.
\displaystyle \therefore 2y=x+z.
\displaystyle \therefore x,y,z\text{ are in A.P.}
\displaystyle \text{Verification: for three terms in A.P., }2y=x+z,\text{ which is satisfied.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{If }S\text{ be the sum, }P\text{ the product and }R\text{ be the sum of the reciprocals}
\displaystyle \text{of }n\text{ terms 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 G.P. be }a,ar,ar^2,\ldots,ar^{n-1}.
\displaystyle S=a\frac{r^n-1}{r-1}.
\displaystyle P=a\cdot ar\cdot ar^2\cdots ar^{n-1} =a^nr^{\frac{n(n-1)}{2}}.
\displaystyle R=\frac{1}{a}+\frac{1}{ar}+\frac{1}{ar^2}+\cdots+\frac{1}{ar^{n-1}}.
\displaystyle R=\frac{1}{a}\frac{r^n-1}{r^{n-1}(r-1)}.
\displaystyle \therefore \frac{S}{R}=a^2r^{n-1}.
\displaystyle \therefore \left(\frac{S}{R}\right)^n =a^{2n}r^{n(n-1)}.
\displaystyle P^2=\left(a^nr^{\frac{n(n-1)}{2}}\right)^2 =a^{2n}r^{n(n-1)}.
\displaystyle \therefore P^2=\left(\frac{S}{R}\right)^n.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 31: }\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 \text{Let }x=2.\overline{357}=2.357357357\ldots
\displaystyle 1000x=2357.357357357\ldots
\displaystyle \text{Subtracting, }1000x-x=2357.357357\ldots-2.357357\ldots
\displaystyle 999x=2355.
\displaystyle \therefore x=\frac{2355}{999}.
\displaystyle \text{Verification: }\frac{2355}{999} =2+\frac{357}{999}=2.\overline{357}.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{If }p^{th},q^{th}\text{ and }r^{th}\text{ terms of an A.P. are in G.P., then the}
\displaystyle \text{common ratio 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 \text{Let the }p^{th},q^{th}\text{ and }r^{th}\text{ terms of the A.P. be }x,y,z.
\displaystyle \text{Let the common difference of the A.P. be }d.
\displaystyle y-x=(q-p)d,\qquad z-y=(r-q)d.
\displaystyle \text{Since }x,y,z\text{ are consecutive terms of a G.P., let its common ratio be }R.
\displaystyle y=Rx,\qquad z=R^2x.
\displaystyle y-x=x(R-1).
\displaystyle z-y=Rx(R-1).
\displaystyle \therefore R=\frac{z-y}{y-x} =\frac{(r-q)d}{(q-p)d}.
\displaystyle R=\frac{r-q}{q-p} =\frac{q-r}{p-q}.
\displaystyle \text{Verification: }\frac{z-y}{y-x}=R\text{ for three consecutive terms of a G.P.}
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{The value of }9^{1/3}\cdot9^{1/9}\cdot9^{1/27}\ldots\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 =9^{\frac13+\frac19+\frac1{27}+\cdots}.
\displaystyle \frac13+\frac19+\frac1{27}+\cdots =\frac{\frac13}{1-\frac13}=\frac12.
\displaystyle \therefore 9^{1/3}\cdot9^{1/9}\cdot9^{1/27}\cdots =9^{1/2}=3.
\displaystyle \text{Verification: the sum of the exponents is }\frac12.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{The sum of an infinite G.P. is }4\text{ and the sum of the cubes}
\displaystyle \text{of its terms is }\frac{64}{7}.\text{ The common ratio of the original G.P. is}
\displaystyle \text{(a) }\frac12\qquad\text{(b) }\frac23\qquad \text{(c) }\frac13\qquad\text{(d) }-\frac12
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and the common ratio be }r.
\displaystyle \frac{a}{1-r}=4.
\displaystyle \therefore a=4(1-r).
\displaystyle \text{The cubes of the terms form an infinite G.P. with first term }a^3
\displaystyle \text{and common ratio }r^3.
\displaystyle \therefore \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}=\frac17.
\displaystyle 7(1-r)^2=1+r+r^2.
\displaystyle 7-14r+7r^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=\frac12\text{ or }r=2.
\displaystyle \text{For an infinite G.P., }|r|<1,\text{ so }r=\frac12.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 35: }\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) }\frac13\qquad \text{(b) }\frac23\qquad \text{(c) }\frac14\qquad \text{(d) }\frac34
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and the common ratio be }r.
\displaystyle a=2(ar+ar^2+ar^3+\cdots).
\displaystyle a=2\left(\frac{ar}{1-r}\right).
\displaystyle 1=\frac{2r}{1-r}.
\displaystyle 1-r=2r\Rightarrow r=\frac13.
\displaystyle \text{The sum of the first two terms is }a+ar=1.
\displaystyle a\left(1+\frac13\right)=1.
\displaystyle \frac{4a}{3}=1\Rightarrow a=\frac34.
\displaystyle \text{Verification: }\frac34+\frac14=1.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{The }n^{th}\text{ term of a G.P. is }128\text{ and the sum of its }n\text{ terms is }255.
\displaystyle \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 a_n=ar^{n-1}=128.
\displaystyle \therefore a2^{n-1}=128.
\displaystyle \therefore a2^n=256.
\displaystyle S_n=\frac{a(2^n-1)}{2-1}=255.
\displaystyle \therefore a(2^n-1)=255.
\displaystyle a2^n-a=255.
\displaystyle 256-a=255.
\displaystyle \therefore a=1.
\displaystyle \text{Verification: }2^{n-1}=128=2^7\Rightarrow n=8.
\displaystyle S_8=1+2+4+\cdots+128=2^8-1=255.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{If second term of a G.P. is }2\text{ and the sum of its infinite terms is }8,
\displaystyle \text{then its first term is}
\displaystyle \text{(a) }\frac14\qquad\text{(b) }\frac12\qquad \text{(c) }2\qquad\text{(d) }4
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and the common ratio be }r.
\displaystyle ar=2\Rightarrow r=\frac{2}{a}.
\displaystyle S_\infty=\frac{a}{1-r}=8.
\displaystyle \frac{a}{1-\frac{2}{a}}=8.
\displaystyle \frac{a^2}{a-2}=8.
\displaystyle a^2-8a+16=0.
\displaystyle (a-4)^2=0.
\displaystyle \therefore a=4.
\displaystyle \text{Verification: }r=\frac12,\quad S_\infty=\frac{4}{1-\frac12}=8.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{If }a,b,c\text{ are in G.P. and }x,y\text{ are AM's between }a,b\text{ and }b,c
\displaystyle \text{respectively, then}
\displaystyle \text{(a) }\frac1x+\frac1y=2\qquad \text{(b) }\frac1x+\frac1y=\frac12
\displaystyle \text{(c) }\frac1x+\frac1y=\frac2a\qquad \text{(d) }\frac1x+\frac1y=\frac2b
\displaystyle \text{Answer:}
\displaystyle x=\frac{a+b}{2},\qquad y=\frac{b+c}{2}.
\displaystyle \therefore \frac1x+\frac1y =\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.
\displaystyle =ab+2b^2+bc=b(a+2b+c).
\displaystyle \therefore \frac1x+\frac1y =\frac{2(a+2b+c)}{b(a+2b+c)}=\frac2b.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{If }A\text{ be one A.M. and }p,q\text{ be two G.M.'s between two numbers,}
\displaystyle \text{then }2A\text{ is equal to}
\displaystyle \text{(a) }\frac{p^3+q^3}{pq}\qquad \text{(b) }\frac{p^3-q^3}{pq}
\displaystyle \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 \text{Let their common ratio be }r.
\displaystyle p=ar,\qquad q=ar^2,\qquad b=ar^3.
\displaystyle 2A=a+b=a+ar^3=a(1+r^3).
\displaystyle \frac{p^3+q^3}{pq} =\frac{a^3r^3+a^3r^6}{a^2r^3}.
\displaystyle =a(1+r^3)=a+b=2A.
\displaystyle \therefore 2A=\frac{p^3+q^3}{pq}.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 40: }\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)
\displaystyle \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=q-p=b-q.
\displaystyle \therefore a=2p-q,\qquad b=2q-p.
\displaystyle \text{Since }G\text{ is the G.M. between }a\text{ and }b,\quad G^2=ab.
\displaystyle \therefore G^2=(2p-q)(2q-p).
\displaystyle \text{Verification: }a,p,q,b\text{ have the same common difference }q-p.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{If }x\text{ is positive, the sum to infinity of the series}
\displaystyle \frac{1}{1+x}-\frac{1-x}{(1+x)^2} +\frac{(1-x)^2}{(1+x)^3}-\frac{(1-x)^3}{(1+x)^4}+\cdots\text{ is}
\displaystyle \text{(a) }\frac12\qquad \text{(b) }\frac34\qquad \text{(c) }1\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{The given series is an infinite G.P. with}
\displaystyle a=\frac{1}{1+x},\qquad r=-\frac{1-x}{1+x}=\frac{x-1}{x+1}.
\displaystyle \text{Since }x>0,\quad \left|\frac{x-1}{x+1}\right|<1.
\displaystyle \therefore S_\infty=\frac{a}{1-r}.
\displaystyle =\frac{\frac{1}{1+x}} {1-\frac{x-1}{x+1}}.
\displaystyle =\frac{\frac{1}{1+x}} {\frac{2}{x+1}}=\frac12.
\displaystyle \text{Verification: }|r|<1\text{ for }x>0,\text{ so the infinite sum exists.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{If }(4^3)(4^6)(4^9)(4^{12})\ldots(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)} =4^{\frac{3x(x+1)}{2}}.
\displaystyle 0.0625=\frac{1}{16}=4^{-2}.
\displaystyle \therefore (0.0625)^{-54} =(4^{-2})^{-54}=4^{108}.
\displaystyle \therefore \frac{3x(x+1)}{2}=108.
\displaystyle x(x+1)=72.
\displaystyle x^2+x-72=0.
\displaystyle (x-8)(x+9)=0.
\displaystyle \therefore x=8\text{ or }x=-9.
\displaystyle \text{Since }x\text{ is the number of factors, }x=8.
\displaystyle \text{Verification: }\frac{3(8)(9)}{2}=108.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 43: }\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} =1+\frac{x}{x+1}+ \left(\frac{x}{x+1}\right)^2+\cdots.
\displaystyle \text{This is an infinite G.P. with } a=1,\qquad r=\frac{x}{x+1}.
\displaystyle \text{Since }x>0,\quad 0<\frac{x}{x+1}<1.
\displaystyle \therefore S_\infty=\frac{a}{1-r} =\frac{1}{1-\frac{x}{x+1}}.
\displaystyle =\frac{1}{\frac{1}{x+1}}=x+1.
\displaystyle \text{Verification: }|r|<1\text{ for }x>0.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{In a G.P. of even number of terms, the sum of all terms is five times}
\displaystyle \text{the sum of the odd terms. The common ratio of the G.P. is}
\displaystyle \text{(a) }-\frac45\qquad\text{(b) }\frac15\qquad \text{(c) }4\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let the G.P. have }2n\text{ terms, first term }a\text{ and common ratio }r.
\displaystyle S=a+ar+ar^2+ar^3+\cdots+ar^{2n-1}.
\displaystyle \text{Let the sum of the odd terms be }S_o.
\displaystyle S_o=a+ar^2+ar^4+\cdots+ar^{2n-2}.
\displaystyle \text{The sum of the even terms is }rS_o.
\displaystyle \therefore S=S_o+rS_o=(1+r)S_o.
\displaystyle \text{Given, }S=5S_o.
\displaystyle \therefore (1+r)S_o=5S_o.
\displaystyle 1+r=5.
\displaystyle \therefore r=4.
\displaystyle \text{Verification: }S=S_o+4S_o=5S_o.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{Let }x\text{ be the A.M. and }y,z\text{ be two G.M.s between two positive}
\displaystyle \text{numbers. Then, }\frac{y^3+z^3}{xyz}\text{ is equal to}
\displaystyle \text{(a) }1\qquad\text{(b) }2\qquad \text{(c) }\frac12\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let the two positive numbers be }a\text{ and }b.
\displaystyle \text{Since }x\text{ is their A.M.,}\quad 2x=a+b.
\displaystyle \text{Since }y,z\text{ are two G.M.s, }a,y,z,b\text{ are in G.P.}
\displaystyle \text{Let the common ratio be }r.
\displaystyle y=ar,\qquad z=ar^2,\qquad b=ar^3.
\displaystyle \therefore 2x=a+ar^3=a(1+r^3).
\displaystyle y^3+z^3=a^3r^3+a^3r^6 =a^3r^3(1+r^3).
\displaystyle xyz=x(ar)(ar^2)=xa^2r^3.
\displaystyle \therefore \frac{y^3+z^3}{xyz} =\frac{a^3r^3(1+r^3)}{xa^2r^3}.
\displaystyle =\frac{a(1+r^3)}{x} =\frac{2x}{x}=2.
\displaystyle \text{Verification: }a(1+r^3)=a+b=2x.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 46: }\text{The product }(32)(32)^{1/6}(32)^{1/36}\ldots\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 =32^{1+\frac16+\frac1{36}+\cdots}.
\displaystyle 1+\frac16+\frac1{36}+\cdots =\frac{1}{1-\frac16}=\frac65.
\displaystyle \therefore \text{Product}=32^{6/5}.
\displaystyle =(2^5)^{6/5}=2^6=64.
\displaystyle \text{Verification: the sum of the exponents is }\frac65.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 47: }\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 two geometric means be }G_1,G_2.
\displaystyle \therefore 1,G_1,G_2,64\text{ are in G.P.}
\displaystyle \text{Let the common ratio be }r.
\displaystyle 64=1\cdot r^3.
\displaystyle \therefore r^3=64=4^3\Rightarrow r=4.
\displaystyle G_1=1\cdot4=4,\qquad G_2=1\cdot4^2=16.
\displaystyle \text{Verification: }1,4,16,64\text{ have common ratio }4.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 48: }\text{In a G.P. if the }(m+n)^{th}\text{ term is }p\text{ and }(m-n)^{th}\text{ term is }q,
\displaystyle \text{then its }m^{th}\text{ term is}
\displaystyle \text{(a) }0\qquad\text{(b) }pq\qquad \text{(c) }\sqrt{pq}\qquad\text{(d) }\frac12(p+q)
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and the common ratio be }r.
\displaystyle ar^{m+n-1}=p,\qquad ar^{m-n-1}=q.
\displaystyle \text{Multiplying the two equations,}
\displaystyle a^2r^{2m-2}=pq.
\displaystyle \left(ar^{m-1}\right)^2=pq.
\displaystyle \therefore ar^{m-1}=\sqrt{pq}.
\displaystyle \text{But }ar^{m-1}\text{ is the }m^{th}\text{ term.}
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 49: }\text{Let }S\text{ be the sum, }P\text{ be the product and }R\text{ be the sum of the}
\displaystyle \text{reciprocals of }3\text{ terms 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})^2:1
\displaystyle \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 }a,ar,ar^2.
\displaystyle S=a+ar+ar^2=a(1+r+r^2).
\displaystyle P=a\cdot ar\cdot ar^2=a^3r^3.
\displaystyle R=\frac1a+\frac1{ar}+\frac1{ar^2}.
\displaystyle R=\frac{1+r+r^2}{ar^2}.
\displaystyle P^2R^3 =(a^3r^3)^2\left(\frac{1+r+r^2}{ar^2}\right)^3.
\displaystyle =a^3(1+r+r^2)^3.
\displaystyle \text{Also, }S^3=a^3(1+r+r^2)^3.
\displaystyle \therefore P^2R^3=S^3.
\displaystyle \therefore P^2R^3:S^3=1:1.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 1: }\text{Write the common difference of an A.P. whose }n^{th}\text{ term is } \\ xn+y.
\displaystyle \text{Answer:}
\displaystyle a_n=xn+y.
\displaystyle a_{n+1}=x(n+1)+y=xn+x+y.
\displaystyle d=a_{n+1}-a_n.
\displaystyle =(xn+x+y)-(xn+y)=x.
\displaystyle \therefore \text{The common difference is }x.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Write the common difference of an A.P., the sum of whose first }n\text{ terms is}
\displaystyle \frac{P}{2}n^2+Qn.
\displaystyle \text{Answer:}
\displaystyle S_n=\frac{P}{2}n^2+Qn.
\displaystyle \text{For an A.P., if }S_n=An^2+Bn,\text{ then the common difference is }2A.
\displaystyle A=\frac{P}{2}.
\displaystyle \therefore d=2\left(\frac{P}{2}\right)=P.
\displaystyle \therefore \text{The common difference is }P.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If the sum of }n\text{ terms of an A.P. is }2n^2+3n,\text{ then write its }n^{th}\text{ term.}
\displaystyle \text{Answer:}
\displaystyle S_n=2n^2+3n.
\displaystyle a_n=S_n-S_{n-1}.
\displaystyle S_{n-1}=2(n-1)^2+3(n-1).
\displaystyle =2n^2-4n+2+3n-3=2n^2-n-1.
\displaystyle a_n=(2n^2+3n)-(2n^2-n-1).
\displaystyle =4n+1.
\displaystyle \therefore \text{The }n^{th}\text{ term is }4n+1.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }\log 2,\ \log(2^x-1)\text{ and }\log(2^x+3)\text{ are in A.P., write the value of }x.
\displaystyle \text{Answer:}
\displaystyle \text{Since the three terms are in A.P.,}
\displaystyle 2\log(2^x-1)=\log 2+\log(2^x+3).
\displaystyle \log(2^x-1)^2=\log\{2(2^x+3)\}.
\displaystyle (2^x-1)^2=2(2^x+3).
\displaystyle \text{Let }2^x=t.
\displaystyle (t-1)^2=2(t+3).
\displaystyle t^2-2t+1=2t+6.
\displaystyle t^2-4t-5=0.
\displaystyle (t-5)(t+1)=0.
\displaystyle t=5\text{ or }t=-1.
\displaystyle \text{Since }t=2^x>0,\quad t=5.
\displaystyle 2^x=5\Rightarrow x=\log_2 5.
\displaystyle \therefore x=\log_2 5.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If the sums of }n\text{ terms of two arithmetic progressions are in the ratio}
\displaystyle 2n+5:3n+4,\text{ then write the ratio of their }m^{th}\text{ terms.}
\displaystyle \text{Answer:}
\displaystyle \frac{S_n}{S'_n}=\frac{2n+5}{3n+4}.
\displaystyle \text{Since }S_n\text{ and }S'_n\text{ are divisible by }n,\text{ we may write}
\displaystyle S_n=kn(2n+5),\qquad S'_n=kn(3n+4).
\displaystyle a_m=S_m-S_{m-1}.
\displaystyle =k\{m(2m+5)-(m-1)[2(m-1)+5]\}.
\displaystyle =k(4m+3).
\displaystyle b_m=S'_m-S'_{m-1}.
\displaystyle =k\{m(3m+4)-(m-1)[3(m-1)+4]\}.
\displaystyle =k(6m+1).
\displaystyle \therefore a_m:b_m=(4m+3):(6m+1).
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Write the sum of first }n\text{ odd natural numbers.}
\displaystyle \text{Answer:}
\displaystyle 1+3+5+\cdots+(2n-1)
\displaystyle \text{Here, }a=1,\quad d=2.
\displaystyle S_n=\frac{n}{2}\{2a+(n-1)d\}.
\displaystyle =\frac{n}{2}\{2+(n-1)2\}.
\displaystyle =\frac{n}{2}(2n)=n^2.
\displaystyle \therefore \text{The sum of first }n\text{ odd natural numbers is }n^2.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Write the sum of first }n\text{ even natural numbers.}
\displaystyle \text{Answer:}
\displaystyle 2+4+6+\cdots+2n
\displaystyle \text{Here, }a=2,\quad d=2.
\displaystyle S_n=\frac{n}{2}\{2a+(n-1)d\}.
\displaystyle =\frac{n}{2}\{4+2(n-1)\}.
\displaystyle =\frac{n}{2}(2n+2)=n(n+1).
\displaystyle \therefore \text{The sum of first }n\text{ even natural numbers is }n(n+1).
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Write the value of }n\text{ for which }n^{th}\text{ terms of the A.P.s }3,10,17,\ldots
\displaystyle \text{and }63,65,67,\ldots\text{ are equal.}
\displaystyle \text{Answer:}
\displaystyle \text{For the first A.P., }a=3,\quad d=7.
\displaystyle T_n=3+(n-1)7=7n-4.
\displaystyle \text{For the second A.P., }a=63,\quad d=2.
\displaystyle T_n=63+(n-1)2=2n+61.
\displaystyle \text{Since the }n^{th}\text{ terms are equal,}
\displaystyle 7n-4=2n+61.
\displaystyle 5n=65\Rightarrow n=13.
\displaystyle \therefore n=13.
\displaystyle \text{The book answer }n=8\text{ does not agree with the printed question.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }\frac{3+5+7+\cdots\text{ up to }n\text{ terms}}{5+8+11+\cdots\text{ up to }10\text{ terms}}=7,
\displaystyle \text{then find the value of }n.
\displaystyle \text{Answer:}
\displaystyle S_n=\frac{n}{2}\{2(3)+(n-1)2\}=n(n+2).
\displaystyle S_{10}=\frac{10}{2}\{2(5)+(10-1)3\}=185.
\displaystyle \frac{n(n+2)}{185}=7.
\displaystyle n(n+2)=1295.
\displaystyle n^2+2n-1295=0.
\displaystyle (n-35)(n+37)=0.
\displaystyle \text{Since }n\text{ is a natural number, }n=35.
\displaystyle \therefore n=35.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }m^{th}\text{ term of an A.P. is }n\text{ and }n^{th}\text{ term is }m,
\displaystyle \text{then write its }p^{th}\text{ term.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and common difference be }d.
\displaystyle a+(m-1)d=n.
\displaystyle a+(n-1)d=m.
\displaystyle \text{Subtracting, }(m-n)d=n-m.
\displaystyle \therefore d=-1.
\displaystyle a+(m-1)(-1)=n.
\displaystyle \therefore a=m+n-1.
\displaystyle T_p=a+(p-1)d.
\displaystyle =(m+n-1)-(p-1)=m+n-p.
\displaystyle \therefore \text{The }p^{th}\text{ term is }m+n-p.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If the sums of }n\text{ terms of two A.P.'s are in the ratio}
\displaystyle (3n+2):(2n+3),\text{ find the ratio of their }12^{th}\text{ terms.}
\displaystyle \text{Answer:}
\displaystyle \frac{S_n}{S'_n}=\frac{3n+2}{2n+3}.
\displaystyle \text{Since the sum of }n\text{ terms of an A.P. contains a factor }n,
\displaystyle S_n=kn(3n+2),\qquad S'_n=kn(2n+3).
\displaystyle T_n=S_n-S_{n-1}.
\displaystyle T_n=k\{n(3n+2)-(n-1)[3(n-1)+2]\}.
\displaystyle =k(6n-1).
\displaystyle T'_n=k\{n(2n+3)-(n-1)[2(n-1)+3]\}.
\displaystyle =k(4n+1).
\displaystyle \therefore \frac{T_{12}}{T'_{12}} =\frac{6(12)-1}{4(12)+1}=\frac{71}{49}.
\displaystyle \therefore \text{Required ratio}=71:49.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If the fifth term of a G.P. is }2,\text{ then write the product of its }9\text{ terms.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and the common ratio be }r.
\displaystyle T_5=ar^4=2.
\displaystyle \text{For a G.P. of }9\text{ terms, the product of terms equidistant from the ends}
\displaystyle \text{is equal to the square of the middle term.}
\displaystyle T_1T_9=T_2T_8=T_3T_7=T_4T_6=T_5^2.
\displaystyle \therefore T_1T_2T_3\cdots T_9=(T_5)^9.
\displaystyle =2^9=512.
\displaystyle \therefore \text{The product of the }9\text{ terms is }512.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }(p+q)^{th}\text{ and }(p-q)^{th}\text{ terms of a G.P. are }m\text{ and }n
\displaystyle \text{respectively, then write its }p^{th}\text{ term.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and the common ratio be }r.
\displaystyle T_{p+q}=ar^{p+q-1}=m.
\displaystyle T_{p-q}=ar^{p-q-1}=n.
\displaystyle \text{Multiplying,}
\displaystyle mn=a^2r^{2p-2}=(ar^{p-1})^2.
\displaystyle mn=(T_p)^2.
\displaystyle \therefore T_p=\sqrt{mn}.
\displaystyle \therefore \text{The }p^{th}\text{ term is }\sqrt{mn}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }\log_x a,\ a^{x/2}\text{ and }\log_b x\text{ are in G.P., then write the value of }x.
\displaystyle \text{Answer:}
\displaystyle \text{Since the three quantities are in G.P., the square of the middle term}
\displaystyle \text{is equal to the product of the other two terms.}
\displaystyle \left(a^{x/2}\right)^2=(\log_x a)(\log_b x).
\displaystyle a^x=\frac{\log a}{\log x}\cdot\frac{\log x}{\log b}.
\displaystyle a^x=\frac{\log a}{\log b}=\log_b a.
\displaystyle \therefore x=\log_a(\log_b a).
\displaystyle \therefore \text{The value of }x\text{ is }\log_a(\log_b a).
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If the sum of an infinite decreasing G.P. is }3\text{ and the sum of the squares}
\displaystyle \text{of its terms is }\frac{9}{2},\text{ then write its first term and common difference.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and the common ratio be }r.
\displaystyle \frac{a}{1-r}=3.
\displaystyle \therefore a=3(1-r).
\displaystyle \text{The squares of the terms form a G.P. with first term }a^2\text{ and common ratio }r^2.
\displaystyle \therefore \frac{a^2}{1-r^2}=\frac{9}{2}.
\displaystyle \frac{9(1-r)^2}{(1-r)(1+r)}=\frac{9}{2}.
\displaystyle \frac{1-r}{1+r}=\frac{1}{2}.
\displaystyle 2-2r=1+r.
\displaystyle \therefore r=\frac{1}{3}.
\displaystyle a=3\left(1-\frac{1}{3}\right)=2.
\displaystyle \therefore \text{The first term is }2\text{ and the common ratio is }\frac{1}{3}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If }p^{th},q^{th}\text{ and }r^{th}\text{ terms of a G.P. are }x,y,z\text{ respectively,}
\displaystyle \text{then write the value of }x^{q-r}y^{r-p}z^{p-q}.
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term be }a\text{ and the common ratio be }R.
\displaystyle x=aR^{p-1},\qquad y=aR^{q-1},\qquad z=aR^{r-1}.
\displaystyle x^{q-r}y^{r-p}z^{p-q}
\displaystyle =(aR^{p-1})^{q-r}(aR^{q-1})^{r-p}(aR^{r-1})^{p-q}.
\displaystyle =a^{q-r+r-p+p-q}R^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)}.
\displaystyle =a^0R^0=1.
\displaystyle \therefore x^{q-r}y^{r-p}z^{p-q}=1.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{If }A_1,A_2\text{ be two AM's and }G_1,G_2\text{ be two GM's between }a\text{ and }b,
\displaystyle \text{then find the value of }\frac{A_1+A_2}{G_1G_2}.
\displaystyle \text{Answer:}
\displaystyle \text{Since }A_1,A_2\text{ are two arithmetic means between }a\text{ and }b,
\displaystyle a,A_1,A_2,b\text{ are in A.P.}
\displaystyle \therefore A_1+A_2=a+b.
\displaystyle \text{Since }G_1,G_2\text{ are two geometric means between }a\text{ and }b,
\displaystyle a,G_1,G_2,b\text{ are in G.P.}
\displaystyle \text{The product of terms equidistant from the ends is equal.}
\displaystyle \therefore G_1G_2=ab.
\displaystyle \therefore \frac{A_1+A_2}{G_1G_2}=\frac{a+b}{ab}.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If second, third and sixth terms of an A.P. are consecutive terms of a G.P.,}
\displaystyle \text{write the common ratio of the G.P.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the first term and common difference of the A.P. be }a\text{ and }d.
\displaystyle T_2=a+d,\qquad T_3=a+2d,\qquad T_6=a+5d.
\displaystyle \text{Since these are consecutive terms of a G.P.,}
\displaystyle (a+2d)^2=(a+d)(a+5d).
\displaystyle a^2+4ad+4d^2=a^2+6ad+5d^2.
\displaystyle d(2a+d)=0.
\displaystyle \text{For a non-constant A.P., }d\ne0,\text{ hence }2a+d=0.
\displaystyle \therefore a=-\frac{d}{2}.
\displaystyle \text{The common ratio of the G.P. is}
\displaystyle r=\frac{a+2d}{a+d} =\frac{-\frac{d}{2}+2d}{-\frac{d}{2}+d} =\frac{\frac{3d}{2}}{\frac{d}{2}}=3.
\displaystyle \therefore \text{The common ratio of the G.P. is }3.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Write the quadratic equation the arithmetic and geometric means of whose}
\displaystyle \text{roots are }A\text{ and }G\text{ respectively.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the roots of the quadratic equation be }\alpha\text{ and }\beta.
\displaystyle A=\frac{\alpha+\beta}{2}\Rightarrow \alpha+\beta=2A.
\displaystyle G=\sqrt{\alpha\beta}\Rightarrow \alpha\beta=G^2.
\displaystyle \text{The quadratic equation with roots }\alpha,\beta\text{ is}
\displaystyle x^2-(\alpha+\beta)x+\alpha\beta=0.
\displaystyle \therefore x^2-2Ax+G^2=0.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Write the product of }n\text{ geometric means between two numbers }a\text{ and }b.
\displaystyle \text{Answer:}
\displaystyle \text{Let }G_1,G_2,\ldots,G_n\text{ be the }n\text{ geometric means between }a\text{ and }b.
\displaystyle \text{Then }a,G_1,G_2,\ldots,G_n,b\text{ are in G.P.}
\displaystyle \text{The product of terms equidistant from the ends is }ab.
\displaystyle \therefore G_1G_2\cdots G_n=(\sqrt{ab})^n.
\displaystyle \therefore G_1G_2\cdots G_n=(ab)^{n/2}.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If }a=1+b+b^2+b^3+\ldots\text{ to }\infty,\text{ then write }b\text{ in terms of }a
\displaystyle \text{given that }|b|<1.
\displaystyle \text{Answer:}
\displaystyle \text{The given infinite series is a G.P. with first term }1\text{ and common ratio }b.
\displaystyle \text{Since }|b|<1,\text{ its sum to infinity is }\frac{1}{1-b}.
\displaystyle \therefore a=\frac{1}{1-b}.
\displaystyle a(1-b)=1.
\displaystyle a-ab=1.
\displaystyle ab=a-1.
\displaystyle \therefore b=\frac{a-1}{a}.
\displaystyle \\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.