JEE Mathematics Sequence and Series MCQs Set 43

Mathematics Objective Questions and Answers: Sequence and Series

Access targeted multiple-choice questions for Sequence and Series designed to align with the latest JEE academic syllabus for JEE Mathematics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

Download Sequence and Series MCQs with Answers

Access the complete set of multiple-choice questions for Sequence and Series below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official JEE textbooks.

Multiple Choice Questions

Question 1: In an A.P. the \(p^{th}\) term is \(q\) and the \((p+q)^{th}\) term is 0. Then the \(q^{th}\) term is
(a) \(-p\)
(b) \(q\)
(c) \(p\)
(d) \(p+q\)
Show Answer & Explanation

Answer: (c) \(p\)

Explanation:
1. \(a+(p-1)d=q\) and \(a+(p+q-1)d=0\); subtracting gives \(qd=-q\), so \(d=-1\).
2. \(a=q+(p-1)=p+q-1\) and \(T_q=a+(q-1)d=p+q-1-(q-1)=p\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Question 2: If \(a,b,c,d,e,f\) are in A.P., then \(e-c\) is equal to
(a) \(2(c-a)\)
(b) \(2(d-e)\)
(c) \((d-c)\)
(d) \(2(d-c)\)
Show Answer & Explanation

Answer: (d) \(2(d-c)\)

Explanation:
1. Let the common difference be \( D \). Then \( e - c = 2D \) and \( d - c = D \).
2. So \( e - c = 2(d - c) \).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Question 3: If 1, \(a\) and \(P\) are in A.P. and 1, \(g\) and \(P\) are in G.P., then
(a) \(1+2a+g^2=0\)
(b) \(1-2a-g^2=0\)
(c) \(1-2a+g^2=0\)
(d) \(1+2a-g^2=0\)
Show Answer & Explanation

Answer: (c) \(1-2a+g^2=0\)

Explanation:
1. \( A.P \).: \(2a=1+P\). \( G.P \).: \(g^2=P\).
2. \(1-2a+g^2=1-(1+P)+P=0\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Question 4: If \(a,b,c\) are in G.P., then \(\dfrac{a-b}{b-c}\) equals
(a) \(ab\)
(b) \(\dfrac ab\text{ or }\dfrac bc\)
(c) \(bc\)
(d) None of these
Show Answer & Explanation

Answer: (b) \(\dfrac ab\text{ or }\dfrac bc\)

Explanation:
1. Let the common ratio be \(r\): \(\dfrac{a-ar}{ar-ar^2}=\dfrac{a(1-r)}{ar(1-r)}=\dfrac1r\).
2. And \(\dfrac ab=\dfrac bc=\dfrac1r\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Question 5: The first and eighth terms of a G.P. are \(x^{-4}\) and \(x^{52}\) respectively. If the second term is \(x^t\), then \(t\) is equal to
(a) 4
(b) -13
(c) \(\dfrac52\)
(d) 3
Show Answer & Explanation

Answer: (a) 4

Explanation:
1. \(r^7=\dfrac{x^{52}}{x^{-4}}=x^{56}\), so \(r=x^8\).
2. Second term \(=x^{-4}\cdot x^8=x^4\), so \(t=4\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Question 6: If the \(p^{th}\) term of an A.P. is \(q\) and its \(q^{th}\) term is \(p\), the common difference is
(a) 0
(b) -1
(c) 2
(d) 1
Show Answer & Explanation

Answer: (b) -1

Explanation:
1. Subtracting the two conditions: \((p-q)d=q-p\), so \(d=-1\) (\( p \) and \( q \) are different).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Question 7: If \(1,x,y,z,16\) are in G.P. with all terms positive, then \(x+y+z\) is
(a) 8
(b) 12
(c) 16
(d) 14
Show Answer & Explanation

Answer: (d) 14

Explanation:
1. \(r^4=16\) so \(r=2\) (positive).
2. \(x+y+z=2+4+8=14\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Question 8: The product of the first nine terms of a G.P. is, in general, equal to
(a) the 9th power of the 5th term
(b) the 9th power of the 4th term
(c) the 4th power of the 9th term
(d) the 5th power of the 9th term
Show Answer & Explanation

Answer: (a) the 9th power of the 5th term

Explanation:
1. Pair terms equidistant from the 5th: their products all equal (5th term)^2.
2. So the product of nine terms is (5th term)^9.

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Question 9: The terms \(\dfrac1{1+\sqrt x},\dfrac1{1-x},\dfrac1{1-\sqrt x}\) of a series are in
(a) G.P.
(b) H.P.
(c) A.P.
(d) A.P. and G.P.
Show Answer & Explanation

Answer: (c) A.P.

Explanation:
1. \(\dfrac1{1+\sqrt x}+\dfrac1{1-\sqrt x}=\dfrac2{1-x}\).
2. So the middle term is the arithmetic mean of the other two: the terms are in \( A.P \).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Question 10: The series \((\sqrt2+1),1,(\sqrt2-1),\ldots\) is in
(a) G.P.
(b) A.P.
(c) H.P.
(d) None of these
Show Answer & Explanation

Answer: (a) G.P.

Explanation:
1. \(\dfrac1{\sqrt2+1}=\sqrt2-1\) and \(\dfrac{\sqrt2-1}1=\sqrt2-1\).
2. The ratio of consecutive terms is constant, so the series is a \( G.P \).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Question 11: Three numbers form an increasing G.P. If the middle number is doubled, the new numbers are in A.P. The common ratio of the G.P. is
(a) \(2-\sqrt3\)
(b) \(2+\sqrt3\)
(c) \(\sqrt3-2\)
(d) \(3+\sqrt2\)
Show Answer & Explanation

Answer: (b) \(2+\sqrt3\)

Explanation:
1. Numbers \(a,ar,ar^2\); \( A.P \). condition \(2(2ar)=a+ar^2\).
2. \(r^2-4r+1=0\) gives \(r=2\pm\sqrt3\); increasing means \(r>1\), so \(r=2+\sqrt3\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Question 12: There are four arithmetic means between 2 and -18. The means are
(a) \(-4,-7,-10,-13\)
(b) \(1,-4,-7,-10\)
(c) \(-2,-6,-10,-14\)
(d) \(-2,-5,-9,-13\)
Show Answer & Explanation

Answer: (c) \(-2,-6,-10,-14\)

Explanation:
1. Six terms in all: \(-18=2+5d\) gives \(d=-4\).
2. The means are \( -2, -6, -10, -14 \).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Question 13: The arithmetic mean of three observations is \(x\). If the values of two observations are \(y\) and \(z\), the third observation is
(a) \(x\)
(b) \(2x-y-z\)
(c) \(y+z-x\)
(d) \(3x-y-z\)
Show Answer & Explanation

Answer: (d) \(3x-y-z\)

Explanation:
1. Sum of the three observations is \(3x\), so the third is \(3x-y-z\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Question 14: If \(\dfrac1{q+r},\dfrac1{r+p},\dfrac1{p+q}\) are in A.P., then
(a) \(p,q,r\text{ are in A.P.}\)
(b) \(\dfrac1p,\dfrac1q,\dfrac1r\text{ are in A.P.}\)
(c) \(p^2,q^2,r^2\text{ are in A.P.}\)
(d) \(p+q+r\text{ is a constant}\)
Show Answer & Explanation

Answer: (c) \(p^2,q^2,r^2\text{ are in A.P.}\)

Explanation:
1. \(\dfrac2{r+p}=\dfrac1{q+r}+\dfrac1{p+q}\) gives \(2(q+r)(p+q)=(r+p)(p+2q+r)\).
2. Expanding and simplifying: \(2q^2=p^2+r^2\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Question 15: If \(G\) is the geometric mean of \(x\) and \(y\), then \(\dfrac1{G^2-x^2}+\dfrac1{G^2-y^2}\) equals
(a) \(G^2\)
(b) \(\dfrac1{G^2}\)
(c) \(\dfrac2{G^2}\)
(d) \(3G^2\)
Show Answer & Explanation

Answer: (b) \(\dfrac1{G^2}\)

Explanation:
1. \(G^2=xy\): \(\dfrac1{x(y-x)}+\dfrac1{y(x-y)}=\dfrac1{y-x}\left(\dfrac1x-\dfrac1y\right)=\dfrac1{xy}\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Question 16: If \(p,q,r\) are in A.P., \(a\) is the G.M. between \(p\) and \(q\) and \(b\) is the G.M. between \(q\) and \(r\), then \(a^2,q^2,b^2\) are in
(a) A.P.
(b) G.P.
(c) H.P.
(d) None of these
Show Answer & Explanation

Answer: (a) A.P.

Explanation:
1. \(a^2=pq\), \(b^2=qr\). Then \(a^2+b^2=q(p+r)=q\cdot2q=2q^2\).
2. So \( q^2 \) is the arithmetic mean of \( a^2 \) and \( b^2 \).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Question 17: The sum of \(n\) terms of the series \(1\cdot3+3\cdot5+5\cdot7+\cdots\) is
(a) \(\dfrac13n(n+1)(2n+1)-n\)
(b) \(\dfrac23n(n+1)(2n+1)-n\)
(c) \(\dfrac32n(n+1)(2n+1)-n\)
(d) \(\dfrac45n(n+1)(2n+1)-n\)
Show Answer & Explanation

Answer: (b) \(\dfrac23n(n+1)(2n+1)-n\)

Explanation:
1. The k-th term is \((2k-1)(2k+1)=4k^2-1\).
2. \(\sum(4k^2-1)=\dfrac{4n(n+1)(2n+1)}6-n=\dfrac23n(n+1)(2n+1)-n\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Question 18: In a geometric progression of positive terms, each term equals the sum of the next two terms. Then the common ratio is
(a) \(\sqrt5\)
(b) \(\dfrac12\left(1-\sqrt5\right)\)
(c) \(\dfrac{\sqrt5}2\)
(d) \(\dfrac12\left(\sqrt5-1\right)\)
Show Answer & Explanation

Answer: (d) \(\dfrac12\left(\sqrt5-1\right)\)

Explanation:
1. \(a=ar+ar^2\) gives \(r^2+r-1=0\).
2. \(r=\dfrac{-1\pm\sqrt5}2\); positive terms need \(r>0\), so \(r=\dfrac{\sqrt5-1}2\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Question 19: The first two terms of a G.P. add up to 12. The sum of the third and the fourth terms is 48. If the terms of the G.P. are alternately positive and negative, then the first term is
(a) -12
(b) -4
(c) 12
(d) 4
Show Answer & Explanation

Answer: (a) -12

Explanation:
1. \(a(1+r)=12\) and \(ar^2(1+r)=48\), so \(r^2=4\).
2. Alternating signs need \( r = -2 \), so \( a(1 - 2) = 12 \) and \( a = -12 \).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Question 20: The harmonic mean of \(\dfrac a{1-ab}\) and \(\dfrac a{1+ab}\) is
(a) \(\dfrac a{1-a^2b^2}\)
(b) \(\dfrac1{1-a^2b^2}\)
(c) \(\dfrac a{1+a^2b^2}\)
(d) \(a\)
Show Answer & Explanation

Answer: (d) \(a\)

Explanation:
1. \(HM=\dfrac{2xy}{x+y}\) with \(xy=\dfrac{a^2}{1-a^2b^2}\) and \(x+y=\dfrac{2a}{1-a^2b^2}\).
2. \(HM=a\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Question 21: The H.M. between the roots of the equation \(x^2-10x+11=0\) is
(a) \(\dfrac{11}5\)
(b) \(\dfrac15\)
(c) \(\dfrac5{21}\)
(d) \(\dfrac{21}{20}\)
Show Answer & Explanation

Answer: (a) \(\dfrac{11}5\)

Explanation:
1. \(\alpha+\beta=10\), \(\alpha\beta=11\).
2. \(HM=\dfrac{2\alpha\beta}{\alpha+\beta}=\dfrac{22}{10}=\dfrac{11}5\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Question 22: If \(m\) arithmetic means are inserted between 1 and 31 so that the ratio of the 7th and \((m-1)^{th}\) means is 5 : 9, then \(m\) is
(a) 10
(b) 14
(c) 11
(d) 12
Show Answer & Explanation

Answer: (b) 14

Explanation:
1. \(d=\dfrac{30}{m+1}\); means are \(1+7d\) and \(1+(m-1)d\).
2. \(\dfrac{m+211}{31m-29}=\dfrac59\) gives \(9m+1899=155m-145\), so \(m=14\).

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

Question 23: Statement I: if a1, a2, ..., an is a sequence, then the expression a1 + a2 + ... + an is called a series. Statement II: those sequences whose terms follow certain patterns are called progressions. Which statements are correct?
(a) Both are true
(b) Only I is false
(c) Only II is false
(d) Both are false
Show Answer & Explanation

Answer: (a) Both are true

Explanation:
1. Both are the standard definitions of a series and of a progression.

Teacher's Note:
1. Express the given condition through the first term and the common difference or ratio, then solve for the unknowns.
2. Remember the standard facts: AP has a constant difference, GP a constant ratio, and HP has reciprocals in AP.

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Sequence and Series Objective Questions & Solutions for JEE Mathematics

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