\displaystyle \textbf{Question 1. }\text{If }7\text{th and }13\text{th terms of an A.P. be }34\text{ and }64\text{ respectively, then its }18\text{th term is}
\displaystyle \text{(a) }87\qquad \text{(b) }88\qquad \text{(c) }89\qquad \text{(d) }90
\displaystyle \text{Answer:}
\displaystyle a+6d=34
\displaystyle a+12d=64
\displaystyle \text{Subtracting, }6d=30
\displaystyle d=5
\displaystyle a+6(5)=34
\displaystyle a=4
\displaystyle \therefore T_{18}=a+17d=4+17(5)=89
\displaystyle \therefore \text{Correct option is (c).}
\\

\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_p=q,\qquad S_q=p
\displaystyle S_n=\frac{n}{2}\{2a+(n-1)d\}
\displaystyle \frac{p}{2}\{2a+(p-1)d\}=q
\displaystyle \frac{q}{2}\{2a+(q-1)d\}=p
\displaystyle \frac{2a+(p-1)d}{2q}=\frac{1}{p},\qquad \frac{2a+(q-1)d}{2p}=\frac{1}{q}
\displaystyle 2a+(p-1)d=\frac{2q}{p},\qquad 2a+(q-1)d=\frac{2p}{q}
\displaystyle \text{Solving gives }2a+(p+q-1)d=-2
\displaystyle \therefore S_{p+q}=\frac{p+q}{2}\{2a+(p+q-1)d\}
\displaystyle =\frac{p+q}{2}(-2)=-(p+q)
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 3. }\text{If the sum of }n\text{ terms of an A.P. be }3n^2-n\text{ and its common difference is }6,
\displaystyle \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 a=S_1=3(1)^2-1=2
\displaystyle \therefore \text{First term}=2
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 4. }\text{Sum of all two digit numbers which when divided by }4\text{ yield unity 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{Required two digit numbers are }13,17,21,\ldots,97.
\displaystyle a=13,\quad d=4,\quad l=97
\displaystyle l=a+(n-1)d
\displaystyle 97=13+(n-1)4
\displaystyle n=22
\displaystyle \therefore \text{Sum}=\frac{22}{2}(13+97)=11\times110=1210
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 5. }\text{In A.M.'s are introduced between }3\text{ and }17\text{ such that the ratio of the last mean}
\displaystyle \text{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 }n\text{ A.M.'s be inserted between }3\text{ and }17.
\displaystyle \text{Then total terms}=n+2
\displaystyle d=\frac{17-3}{n+1}=\frac{14}{n+1}
\displaystyle \text{First mean}=3+d,\qquad \text{last mean}=17-d
\displaystyle \frac{17-d}{3+d}=3
\displaystyle 17-d=9+3d
\displaystyle 4d=8
\displaystyle d=2
\displaystyle \frac{14}{n+1}=2
\displaystyle n+1=7
\displaystyle n=6
\displaystyle \therefore \text{Correct option is (a).}
\\

\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 } \\ \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}
\displaystyle \text{This implies }S_n=kn^2\text{ for some constant }k.
\displaystyle a_n=S_n-S_{n-1}=kn^2-k(n-1)^2
\displaystyle =k(2n-1)
\displaystyle \therefore a_m=k(2m-1),\qquad a_n=k(2n-1)
\displaystyle \therefore \frac{a_m}{a_n}=\frac{2m-1}{2n-1}
\displaystyle \therefore \text{Correct option is (b).}
\\

\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 terms is }36,
\displaystyle \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 S_n=\frac{n}{2}(a+l)
\displaystyle 36=\frac{n}{2}(1+11)
\displaystyle 36=6n
\displaystyle n=6
\displaystyle \therefore \text{Correct option is (b).}
\\

\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 terms is }164?
\displaystyle \text{(a) }26\text{th}\qquad \text{(b) }27\text{th}\qquad \text{(c) }28\text{th}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle S_n=3n^2+5n
\displaystyle S_{n-1}=3(n-1)^2+5(n-1)
\displaystyle =3n^2-n-2
\displaystyle a_n=S_n-S_{n-1}
\displaystyle =(3n^2+5n)-(3n^2-n-2)
\displaystyle =6n+2
\displaystyle 6n+2=164
\displaystyle 6n=162
\displaystyle n=27
\displaystyle \therefore 164\text{ is the }27\text{th term.}
\displaystyle \therefore \text{Correct option is (b).}
\\

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

\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 the sum of the series}
\displaystyle \sin d\left[\mathrm{cosec}\ a_1\mathrm{cosec}\ a_2+\mathrm{cosec}\ a_2\mathrm{cosec}\ a_3+\cdots+\mathrm{cosec}\ a_{n-1}\mathrm{cosec}\ a_n\right]\text{ is}
\displaystyle \text{(a) }\sec a_1-\sec a_n\qquad \text{(b) }\mathrm{cosec}\ a_1-\mathrm{cosec}\ a_n\qquad \text{(c) }\cot a_1-\cot a_n\qquad \text{(d) }\tan a_1-\tan a_n
\displaystyle \text{Answer:}
\displaystyle a_{r+1}-a_r=d
\displaystyle \frac{\sin d}{\sin a_r\sin 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{Required sum}=(\cot a_1-\cot a_2)+(\cot a_2-\cot a_3)+\cdots+(\cot a_{n-1}-\cot a_n)
\displaystyle =\cot a_1-\cot a_n
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 11. }\text{In the arithmetic progression whose common difference is non-zero,} \\ \text{the sum of first }3n\text{ terms} \text{ is equal to the sum of next }n\text{ terms. Then the ratio of the} \\ \text{sum of the first }2n\text{ terms to the next }2n\text{ terms is}
\displaystyle \text{(a) }\frac{1}{5}\qquad \text{(b) }\frac{2}{3}\qquad \text{(c) }\frac{3}{4}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle S_{3n}=S_{4n}-S_{3n}
\displaystyle 2S_{3n}=S_{4n}
\displaystyle 2\cdot\frac{3n}{2}\{2a+(3n-1)d\}=\frac{4n}{2}\{2a+(4n-1)d\}
\displaystyle 3\{2a+(3n-1)d\}=2\{2a+(4n-1)d\}
\displaystyle 2a+(n-1)d=0
\displaystyle S_{2n}=\frac{2n}{2}\{2a+(2n-1)d\}=n\{nd\}=n^2d
\displaystyle S_{4n}-S_{2n}=2n\{2a+(4n-1)d\}-n^2d
\displaystyle =2n\{3nd\}-n^2d=5n^2d
\displaystyle \therefore \text{Required ratio}=n^2d:5n^2d=1:5
\displaystyle \therefore \text{Correct option is (a).}
\\

\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\left[\sec a_1\sec a_2+\sec a_2\sec a_3+\cdots+\sec a_{n-1}\sec a_n\right]\text{ is}
\displaystyle \text{(a) }\sec a_1-\sec a_n\qquad \text{(b) }\mathrm{cosec}\ a_1-\mathrm{cosec}\ a_n\qquad \text{(c) }\cot a_1-\cot a_n\qquad \text{(d) }\tan a_n-\tan a_1
\displaystyle \text{Answer:}
\displaystyle a_{r+1}-a_r=d
\displaystyle \sin d\sec a_r\sec a_{r+1}
\displaystyle =\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{Required sum}=(\tan a_2-\tan a_1)+(\tan a_3-\tan a_2)+\cdots+(\tan a_n-\tan a_{n-1})
\displaystyle =\tan a_n-\tan a_1
\displaystyle \therefore \text{Correct option is (d).}
\\

\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 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 \text{Their sum is }4a=50
\displaystyle a=\frac{25}{2}
\displaystyle \text{Greatest number}=4\times\text{least number}
\displaystyle a+3d=4(a-3d)
\displaystyle 15d=3a
\displaystyle d=\frac{a}{5}=\frac{5}{2}
\displaystyle \therefore \text{Numbers are }5,\ 10,\ 15,\ 20
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 14. }\text{If arithmetic means are inserted between }1\text{ and }31\text{ such that the ratio of}
\displaystyle \text{the first mean and }n\text{th 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 \text{First mean}=1+d
\displaystyle n\text{th mean}=1+nd
\displaystyle \frac{1+d}{1+nd}=\frac{3}{29}
\displaystyle 29+29d=3+3nd
\displaystyle 26=d(3n-29)
\displaystyle \text{Also, the last term is }31.
\displaystyle 31=1+(n+1)d
\displaystyle d=\frac{30}{n+1}
\displaystyle 26=\frac{30}{n+1}(3n-29)
\displaystyle 13(n+1)=15(3n-29)
\displaystyle 13n+13=45n-435
\displaystyle 32n=448
\displaystyle n=14
\displaystyle \therefore \text{Correct option is (d).}
\\

\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 S_n=\frac{n}{2}[2a+(n-1)d]
\displaystyle S_{n-1}=\frac{n-1}{2}[2a+(n-2)d]
\displaystyle S_{n-2}=\frac{n-2}{2}[2a+(n-3)d]
\displaystyle \text{Using }d=S_n-kS_{n-1}+S_{n-2}
\displaystyle \text{and simplifying, coefficient of }a\text{ must be zero.}
\displaystyle n-k(n-1)+(n-2)=0
\displaystyle 2n-2-k(n-1)=0
\displaystyle (n-1)(2-k)=0
\displaystyle \therefore k=2
\displaystyle \therefore \text{Correct option is (b).}
\\

\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 S=\frac{n}{2}(a+l)
\displaystyle l-a=(n-1)d
\displaystyle d=\frac{l-a}{n-1}
\displaystyle =\frac{(l-a)(l+a)}{(n-1)(l+a)}
\displaystyle =\frac{l^2-a^2}{2S-(l+a)}
\displaystyle \therefore k=2S
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 17. }\text{If the sum of first }n\text{ even natural numbers is equal to }k\text{ times the sum of}
\displaystyle \text{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\cdot n^2
\displaystyle k=\frac{n+1}{n}
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 18. }\text{If the first, second and last term of an A.P. are }a,b\text{ and }2a\text{ respectively,}
\displaystyle \text{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{First term}=a,\qquad \text{second term}=b
\displaystyle \therefore d=b-a
\displaystyle \text{Last term}=2a
\displaystyle 2a=a+(n-1)(b-a)
\displaystyle n-1=\frac{a}{b-a}
\displaystyle n=\frac{b}{b-a}
\displaystyle S=\frac{n}{2}(a+2a)
\displaystyle =\frac{3an}{2}
\displaystyle =\frac{3a}{2}\cdot\frac{b}{b-a}
\displaystyle =\frac{3ab}{2(b-a)}
\displaystyle \therefore \text{Correct option is (c).}
\\

\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 A.P. have }n\text{ terms, where }n\text{ is odd.}
\displaystyle S_1=\frac{n}{2}\{2a+(n-1)d\}
\displaystyle \text{Number of odd placed terms}=\frac{n+1}{2}
\displaystyle \text{Odd placed terms form an A.P. with first term }a\text{ and common difference }2d.
\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{Correct option is (a).}
\\

\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 the sum of }r\text{ terms}
\displaystyle \text{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 S_n=n^2p,\qquad S_m=m^2p
\displaystyle \text{This implies }S_r=r^2p.
\displaystyle \therefore S_p=p^2\cdot p=p^3
\displaystyle \therefore \text{Correct option is (c).}
\\

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

\displaystyle \textbf{Question 22. }\text{The }10\text{th common term between the A.P.s }3,7,11,15,\ldots\text{ and }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{First A.P. has terms }3+4r.
\displaystyle \text{Second A.P. has terms }1+5s.
\displaystyle 3+4r=1+5s
\displaystyle 4r-5s=-2
\displaystyle \text{The common terms form an A.P. with first common term }11\text{ and common difference }\mathrm{lcm}(4,5)=20.
\displaystyle \therefore 10\text{th common term}=11+(10-1)20
\displaystyle =191
\displaystyle \therefore \text{Correct option is (a).}
\\

\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 the sum of }
\displaystyle r\text{ terms } \text{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 S_n=n^2q,\qquad S_m=m^2q
\displaystyle \text{This implies }S_r=r^2q.
\displaystyle \therefore S_q=q^2\cdot q=q^3
\displaystyle \therefore \text{Correct option is (c).}
\\

\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,\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 S_n=\frac{n}{2}\{2a+(n-1)d\}
\displaystyle S_{2n}=n\{2a+(2n-1)d\}
\displaystyle S_{2n}=3S_n
\displaystyle n\{2a+(2n-1)d\}=\frac{3n}{2}\{2a+(n-1)d\}
\displaystyle 2\{2a+(2n-1)d\}=3\{2a+(n-1)d\}
\displaystyle 2a-(n+1)d=0
\displaystyle \therefore 2a=(n+1)d
\displaystyle S_{3n}=\frac{3n}{2}\{2a+(3n-1)d\}
\displaystyle =\frac{3n}{2}\{(n+1)d+(3n-1)d\}
\displaystyle =\frac{3n}{2}(4nd)=6n^2d
\displaystyle S_n=\frac{n}{2}\{(n+1)d+(n-1)d\}=n^2d
\displaystyle \therefore S_{3n}:S_n=6:1
\displaystyle \therefore \text{Correct option is (b).}
\\

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