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Sequences and Series questions for IPMAT, with solutions

10 free Sequences and Series practice questions for IPMAT Indore, Rohtak and JIPMAT, 5 from past papers, each with the answer and a full worked solution. No sign-in. Try each one first, then open the solution. The lessons for this chapter are in Sequences and Series on Learn.

1. Question 1Multiple choiceAsked in IPMAT Rohtak 2019Arithmetic Progression

   The sum of third and ninth term of an A.P is 8. Find the sum of the first 11 terms of the progression.

   1. A.44
   2. B.22
   3. C.19
   4. D.None of the above
   Show the answer and solution

   Answer: A. 44

   $T_3+T_9=(a+2d)+(a+8d)=2a+10d=8$, so $a+5d=4$, i.e. $T_6=4$. The sum of the first $11$ terms is $S_{11}=\frac{11}{2}(2a+10d)=11(a+5d)$. So $S_{11}=11\cdot4=44$.

2. Question 2Multiple choiceAsked in JIPMAT 2023Arithmetic Progression

   Assertion (A): The sum of the first hundred even natural numbers divisible by $5$ is $45050$. Reason (R): The sum of the first $n$ terms of an arithmetic progression is $S_n=\frac{n}{2}(a+l)$, where $a$ is the first term and $l$ the last term.

   1. A.Both (A) and (R) are true and (R) is the correct explanation of (A).
   2. B.Both (A) and (R) are true but (R) is NOT the correct explanation of (A).
   3. C.(A) is true but (R) is false.
   4. D.(A) is false but (R) is true.
   Show the answer and solution

   Answer: D. (A) is false but (R) is true.

   An even number divisible by $5$ is a multiple of $10$. The first hundred are $10,20,\ldots,1000$. Their sum is $\frac{100}{2}(10+1000)=50\cdot1010=50500$, not $45050$. So the Assertion is false. The Reason, $S_n=\frac n2(a+l)$, is the standard AP sum formula and is true.

3. Question 3Multiple choiceAsked in JIPMAT 2024Geometric Progression

   Assertion (A): The sum of $n$ terms of the progression $1+\frac12+\frac1{2^2}+\frac1{2^3}+\cdots$ is $\frac{2^{n-1}-1}{2^{n-1}}$. Reason (R): The sum of a geometric series having $n$ terms is $S_n=\frac{a(1-r^n)}{1-r}$, where $a$ is the first term and $r$ is the common ratio.

   1. A.Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of (A).
   2. B.Both Assertion (A) and Reason (R) are true but Reason (R) is NOT the correct explanation of (A).
   3. C.(A) is true but (R) is false.
   4. D.(A) is false but (R) is true.
   Show the answer and solution

   Answer: D. (A) is false but (R) is true.

   For $1+\frac12+\frac14+\cdots$ ($n$ terms), $a=1$ and $r=\frac12$. By the formula, $S_n=\frac{1-(1/2)^n}{1-1/2}=2-\frac{1}{2^{n-1}}=\frac{2^n-1}{2^{n-1}}$. The assertion says $\frac{2^{n-1}-1}{2^{n-1}}$, which is wrong (for $n=1$ it gives $0$, but the sum is $1$). So the Assertion is false and the Reason (the standard formula) is true.

4. Question 4Multiple choiceWritten by IPMATwayAGP, HP and Means

   The $2$nd term of a harmonic progression is $\dfrac15$ and the $5$th term is $\dfrac1{14}$. Which term of the progression is $\dfrac1{32}$?

   1. A.$12$
   2. B.$11$
   3. C.$32$
   4. D.$10$
   Show the answer and solution

   Answer: B. $11$

   The reciprocals form an AP with $2$nd term $5$ and $5$th term $14$. So $a+d=5$ and $a+4d=14$, giving $d=3$ and $a=2$. We need $2+3(k-1)=32$, so $k-1=10$ and $k=11$. The $11$th term of the HP is $\frac{1}{32}$.

5. Question 5Multiple choiceWritten by IPMATwayAGP, HP and Means

   If the positive numbers $a,b,c,d$ are in arithmetic progression, then $abc,\ abd,\ acd,\ bcd$ are in

   1. A.HP
   2. B.none of these
   3. C.GP
   4. D.AP
   Show the answer and solution

   Answer: A. HP

   Each product leaves out one of the four numbers: $abc=\frac{abcd}{d}$, $abd=\frac{abcd}{c}$, $acd=\frac{abcd}{b}$, $bcd=\frac{abcd}{a}$. Their reciprocals are $\frac{d}{abcd},\frac{c}{abcd},\frac{b}{abcd},\frac{a}{abcd}$. Since $d,c,b,a$ is an AP, dividing by the constant $abcd$ keeps it an AP. Numbers whose reciprocals are in AP are in HP.

6. Question 6Multiple choiceAsked in IPMAT Indore 2021Arithmetic Progression

   Let $S_n$ be the sum of the first $n$ terms of an A.P. $\{a_n\}$. If $S_5=S_9$, what is the ratio $a_3:a_5$?

   1. A.9:5
   2. B.5:9
   3. C.3:5
   4. D.5:3
   Show the answer and solution

   Answer: A. 9:5

   $S_5=S_9$ means $a_6+a_7+a_8+a_9=0$, i.e. $4a+26d=0$, so $a=-\frac{13}{2}d$. Then $a_3=a+2d=-\frac92d$ and $a_5=a+4d=-\frac52d$. So $a_3:a_5=\frac92:\frac52=9:5$.

7. Question 7Multiple choiceAsked in IPMAT Indore 2023Geometric Progression

   A person standing at the center of an open ground first walks $32$ meters towards the east, takes a right turn and walks $16$ meters, takes another right turn and walks $8$ meters, and so on. How far will the person be from the original starting point after an infinite number of such walks in this pattern?

   1. A.$\frac{64}{\sqrt{5}}$ meters
   2. B.$64$ meters
   3. C.$32$ meters
   4. D.$\frac{32}{\sqrt{5}}$ meters
   Show the answer and solution

   Answer: A. $\frac{64}{\sqrt{5}}$ meters

   The legs go E, S, W, N, E, ... with lengths $32,16,8,4,\dots$. The east-west legs $32,-8,2,\dots$ form a GP with ratio $-\frac{1}{4}$: sum $\frac{32}{5/4}=25.6$. The north-south legs $-16,4,-1,\dots$ sum to $-\frac{16}{5/4}=-12.8$. Distance $=\sqrt{25.6^2+12.8^2}=12.8\sqrt{5}=\frac{64}{\sqrt{5}}$ m.

8. Question 8Multiple choiceWritten by IPMATwaySpecial Series

   The sum of the first $n$ terms of a sequence is $S_n=2n^2+5n+1$ for every $n\ge1$. Then $t_1+t_{10}$ equals

   1. A.$51$
   2. B.$55$
   3. C.$251$
   4. D.$50$
   Show the answer and solution

   Answer: A. $51$

   $t_1=S_1=2+5+1=8$. For $n\ge2$, $t_n=S_n-S_{n-1}$. $S_{10}=200+50+1=251$ and $S_9=162+45+1=208$, so $t_{10}=251-208=43$ (the general form is $t_n=4n+3$ for $n\ge2$). Note $t_1=8$, not $4\cdot1+3=7$, because of the constant term in $S_n$. So $t_1+t_{10}=8+43=51$.

9. Question 9Type the answerWritten by IPMATwayAGP, HP and Means

   Find $1\cdot2+2\cdot2^2+3\cdot2^3+\dots+10\cdot2^{10}$.

   Show the answer and solution

   Answer: 18434

   Let $S=1\cdot2+2\cdot2^2+\cdots+10\cdot2^{10}$. Then $2S=1\cdot2^2+\cdots+9\cdot2^{10}+10\cdot2^{11}$. Subtract: $S=2S-S=10\cdot2^{11}-(2+2^2+\cdots+2^{10})$. The GP part is $2^{11}-2=2046$. So $S=20480-2046=18434$.

10. Question 10Multiple choiceWritten by IPMATwaySpecial Series

    If $a_1,a_2,\dots,a_n$ are positive numbers in arithmetic progression, then $\dfrac{1}{\sqrt{a_1}+\sqrt{a_2}}+\dfrac{1}{\sqrt{a_2}+\sqrt{a_3}}+\dots+\dfrac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}$ equals

    1. A.$\dfrac{n-1}{\sqrt{a_1}-\sqrt{a_n}}$
    2. B.$\dfrac{n-1}{\sqrt{a_n}-\sqrt{a_1}}$
    3. C.$\dfrac{n-1}{\sqrt{a_1}+\sqrt{a_n}}$
    4. D.$\dfrac{n}{\sqrt{a_1}+\sqrt{a_n}}$
    Show the answer and solution

    Answer: C. $\dfrac{n-1}{\sqrt{a_1}+\sqrt{a_n}}$

    Rationalise one term: $\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}}=\frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{a_{k+1}-a_k}=\frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d}$. Adding the $n-1$ terms, the numerators telescope to $\sqrt{a_n}-\sqrt{a_1}$. So the sum is $\frac{\sqrt{a_n}-\sqrt{a_1}}{d}=\frac{a_n-a_1}{d(\sqrt{a_n}+\sqrt{a_1})}$. Since $a_n-a_1=(n-1)d$, the sum is $\frac{n-1}{\sqrt{a_1}+\sqrt{a_n}}$.