Chapter 9

71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers · 120 exercises

Problem 1

The common difference of the A.P. \(b_{1}, b_{2}, \ldots, b_{m}\) is 2 more than the common difference of A.P. \(a_{1}, a_{2}, \ldots, a_{n}\). If \(\mathrm{a}_{40}=-159, \mathrm{a}_{100}=-399\) and \(\mathrm{b}_{100}=\mathrm{a}_{70}\), then \(\mathrm{b}_{1}\) is equal to: [Sep. 06, 2020 (II)] (a) 81 (b) \(-127\) (c) \(-81\) (d) 127

5 step solution

Problem 2

(a) 81 (b) \(-127\) (c) \(-81\) (d) 127 If \(3^{2 \sin 2 \alpha-1}, 14\) and \(3^{4-2 \sin 2 \alpha}\) are the first three terms of an A.P. for some \(\alpha\), then the sixth term of this A.P is: [Sep. \(05,2020(\mathrm{I})]\) (a) 66 (b) 81 (c) 65 (d) 78

6 step solution

Problem 3

If the sum of the first 20 terms of the series \(\log _{\left(7^{1 / 2}\right)} x+\log _{\left(7^{1 / 3}\right)} x+\log _{\left(7^{1 / 4}\right)} x+\ldots\) is 460 , then \(x\) is equal to : [Sep. 05, 2020 (II)] (a) \(7^{2}\) (b) \(7^{1 / 2}\) (c) \(e^{2}\) (d) \(7^{46 / 21}\)

6 step solution

Problem 4

Let \(a_{1}, a_{2}, \ldots ., a_{n}\) be a given A.P. whose common difference is an integer and \(S_{n}=a_{1}+a_{2}+\ldots+a_{n} .\) If \(a_{1}=1, a_{n}=300\) and \(15 \leq n \leq 50\), then the ordered pair \(\left(S_{n-4}, a_{n-4}\right)\) is equal to : (a) \((2490,249)\) (b) \((2480,249)\) (c) \((2480,248)\) (d) \((2490,248)\)

7 step solution

Problem 5

If the first term of an A.P. is 3 and the sum of its first 25 terms is equal to the sum of its next 15 terms, then the common difference of this A.P. is: \(\quad\) [Sep. 03, 2020 (I)] (a) \(\frac{1}{6}\) (b) \(\frac{1}{5}\) (c) \(\frac{1}{4}\) (d) \(\frac{1}{7}\)

7 step solution

Problem 7

If the sum of first 11 terms of an A.P., \(a_{1}, a_{2}, a_{3}, \ldots\) is 0 \(\left(a_{1} \neq 0\right)\), then the sum of the A.P., \(a_{1}, a_{3}, a_{5}, \ldots, a_{23}\) is \(k a_{1}\), where \(k\) is equal to : [Sep. 02, 2020 (II)] (a) \(-\frac{121}{10}\) (b) \(\frac{121}{10}\) (c) \(\frac{72}{5}\) (d) \(-\frac{72}{5}\)

6 step solution

Problem 8

The number of terms common to the two A.P's \(3,7,11, \ldots\), 407 and \(2,9,16, \ldots, 709\) is \(\quad\) [NAJan. 9, 2020 (II)]

8 step solution

Problem 9

If the \(10^{\text {th }}\) term of an A.P. is \(\frac{1}{20}\) and its \(20^{\text {th }}\) term is \(\frac{1}{10}\), then the sum of its first 200 terms is: \(\quad\) [Jan. 8, \(\mathbf{2 0 2 0}\) (II)] (a) 50 (b) \(50 \frac{1}{4}\) (c) 100 (d) \(100 \frac{1}{2}\)

8 step solution

Problem 10

Let \(f: R \rightarrow R\) be such that for all \(x \in R,\left(2^{1+x}+2^{1-x}\right), f(x)\) and \(\left(3^{x}+3^{-x}\right)\) are in A.P., then the minimum value of \(f(x)\) is: [Jan. 8, 2020 (I)] (a) 2 (b) 3 (c) \(\underline{0}\) (d) 4

6 step solution

Problem 11

Five numbers are in A.P., whose sum is 25 and product is 2520 . If one of these five numbers is \(-\frac{1}{2}\), then the greatest number amongst them is: [Jan. 7, 2020 (I)] (a) 27 (b) 7 (c) \(\frac{21}{2}\) (d) 16

5 step solution

Problem 12

Let \(\mathrm{S}_{\mathrm{n}}\) denote the sum of the first \(n\) terms of an A.P. If \(\mathrm{S}_{4}=16\) and \(S_{6}=-48\), then \(S_{10}\) is equal to : [April 12, 2019 (I)\\} (a) \(-260\) (b) \(-410\) (c) \(-320\) (d) \(-380\)

4 step solution

Problem 13

If \(a_{1}, a_{2}, a_{3}, \ldots \ldots\) are in A.P. such that \(a_{1}+a_{7}+a_{16}=40\), then the sum of the first 15 terms of this A.P. is : [April 12, 2019 (II)] (a) 200 (b) 280 (c) 120 (d) 150

6 step solution

Problem 14

If \(a_{1}, a_{2}, a_{3}, \ldots . . a_{n}\) are in A.P. and \(a_{1}+a_{4}+a_{7}+\ldots+a_{16}=114\) then \(a_{1}+a_{6}+a_{11}+a_{16}\) is equal to : \(\quad\) April 10,2019 (I)] (a) 98 (b) 76 (c) 38 (d) 64

7 step solution

Problem 15

Let the sum of the first \(\mathrm{n}\) terms of a non-constant A.P., \(a_{1}, a_{2}, a_{3}, \ldots \ldots \ldots \ldots . .\) be \(50 n+\frac{n(n-7)}{2} A\), where \(A\) is a constant. If \(\mathrm{d}\) is the common difference of this A.P., then the ordered pair \(\left(\mathrm{d}, \mathrm{a}_{50}\right)\) is equal to: \(\quad\) [April \(\mathbf{0 9}, \mathbf{2 0 1 9}\) (I)] (a) \((50,50+46 \mathrm{~A})\) (b) \((50,50+45 \mathrm{~A})\) (c) \((\mathrm{A}, 50+45 \mathrm{~A})\) (d) \((\mathrm{A}, 50+46 \mathrm{~A})\)

5 step solution

Problem 16

Let \(\sum_{\mathrm{k}=1}^{10} \mathrm{f}(\mathrm{a}+\mathrm{k})=16\left(2^{10}-1\right)\), where the function \(\mathrm{f}\) satisfies \(f(x+y)=f(x) f(y)\) for all natural numbers \(x, y\) and \(f(a)=2\). Then the natural number ' \(\mathrm{a}\) ' is: [April 09, 2019 (I)] (a) 2 (b) 16 (c) 4 (d) 3

6 step solution

Problem 17

If the sum and product of the first three terms in an A.P. are 33 and 1155 , respectively, then a value of its \(11^{\text {th }}\) term is: \(\quad\) [April09,2019 (II)] (a) \(-35\) (b) 25 (c) \(-36\) (d) \(-25\)

9 step solution

Problem 18

The sum of all natural numbers ' \(n\) ' such that \(1001\) is: [April 08, 2019 (I)] (a) 3203 (b) 3303 (c) 3221 (d) 3121

7 step solution

Problem 19

If \({ }^{n} C_{4},{ }^{n} C_{5}\) and \({ }^{n} C_{6}\) are in A.P., then \(n\) can be : [Jan.12, 2019 (II)] (a) 9 (b) 14 (c) 11 (d) 12

6 step solution

Problem 20

If 19 th term of a non-zero A.P. is zero, then its (49th term): ( 29 th term ) is: [Jan. 11, 2019 (II)] (a) \(4: 1\) (b) \(1: 3\) (c) \(3: 1\) (d) \(2: 1\)

6 step solution

Problem 21

The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is: [Jan. 10, 2019 (I)] (a) 1256 (b) 1465 (c) 1365 (d) 1356

7 step solution

Problem 22

Let \(a_{1}, a_{2}, \ldots ., a_{30}\) be an A.P., \(S=\sum_{i=1}^{30} a_{i}\) and \(\mathrm{T}=\sum_{i=1}^{15} \mathrm{a}_{(2 i-1)^{*}}\) If \(\mathrm{a}_{5}=27\) and \(\mathrm{S}-2 \mathrm{~T}=75\), then \(\mathrm{a}_{10}\) is equal to: [Jan. 09, 2019 (I)] (a) 52 (b) 57 (c) 47 (d) 42

10 step solution

Problem 25

Let \(a_{1}, a_{2}, a_{3}, \ldots, a_{49}\) be in A.P. such that \(\sum_{\mathrm{k}=0}^{12} \mathrm{a}_{4 \mathrm{k}+1}=416\) and \(\mathrm{a}_{9}+\mathrm{a}_{43}=66 .\) If \(\mathrm{a}_{1}^{2}+\mathrm{a}_{2}^{2}+\ldots+\mathrm{a}_{17}^{2}=140 \mathrm{~m}\), then \(\mathrm{m}\) is equal to: (a) 68 (b) 34 (c) 33 (d) 66

8 step solution

Problem 26

For any three positive real numbers \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\), \(9\left(25 a^{2}+b^{2}\right)+25\left(c^{2}-3 a c\right)=15 b(3 a+c)\). Then (a) \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\) are in G.P. (b) \(\mathrm{b}, \mathrm{c}\) and a are in G.P. (c) \(\mathrm{b}, \mathrm{c}\) and \(\mathrm{a}\) are in A.P. (d) \(\mathrm{a}, \mathrm{b}\) and \(\mathrm{c}\) are in A.P.

7 step solution

Problem 27

If three positive numbers \(a, b\) and \(c\) are in A.P. such that \(\mathrm{abc}=8\), then the minimum possible value of \(b\) is : [OnlineApril9, 2017] (a) 2 (b) \(4^{\frac{1}{3}}\) (c) \(4^{\frac{2}{3}}\) (d) 4

5 step solution

Problem 28

Let \(a_{1}, a_{2}, a_{3}, \ldots, a_{n}\), be in A.P. If \(a_{3}+a_{7}+a_{11}+a_{15}=72\), then the sum of its first 17 terms is equal to: [Online April 10, 2016] (a) 306 (b) 204 (c) 153 (d) 612

6 step solution

Problem 29

Let \(\alpha\) and \(\beta\) be the roots of equation \(\mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}=0, \mathrm{p} \neq 0\). If \(\mathrm{p}, \mathrm{q}, \mathrm{r}\) are in A.P and \(\frac{1}{\alpha}+\frac{1}{\beta}=4\), then the value of \(|\alpha-\beta|\) is: (a) \(\frac{\sqrt{34}}{9}\) (b) \(\frac{2 \sqrt{13}}{9}\) (c) \(\frac{\sqrt{61}}{9}\) (d) \(\frac{2 \sqrt{17}}{9}\)

6 step solution

Problem 30

The sum of the first 20 terms common between the series 3 \(+7+11+15+\ldots \ldots . . .\) and \(1+6+11+16+\ldots \ldots\), is |Online April 11, 2014] (a) 4000 (b) 4020 (c) 4200 (d) 4220

8 step solution

Problem 31

Given an A.P. whose terms are all positive integers. The sum of its first nine terms is greater than 200 and less than 220 . If the second term in it is 12 , then its \(4^{\text {th }}\) term is: |Online April 9, 2014] (a) 8 (b) 16 (c) 20 (d) 24

8 step solution

Problem 32

If \(a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots .\) are in A.P. such that \(a_{4}-a_{7}+a_{10}=m\), then the sum of first 13 terms of this A.P., is: [Online April 23, 2013] (a) \(10 \mathrm{~m}\) (b) \(12 \mathrm{~m}\) (c) \(13 \mathrm{~m}\) (d) \(15 \mathrm{~m}\)

6 step solution

Problem 33

Given sum of the first \(n\) terms of an A.P. is \(2 n+3 n^{2}\). Another A.P. is formed with the same first term and double of the common difference, the sum of \(n\) terms of the new A.P. is : [Online April 22, 2013] (a) \(n+4 n^{2}\) (b) \(6 n^{2}-n\) (c) \(n^{2}+4 n\) (d) \(3 n+2 n^{2}\)

6 step solution

Problem 34

Let \(a_{1}, a_{2}, a_{3}, \ldots\) be an A.P, such that \(\frac{a_{1}+a_{2}+\ldots+a_{p}}{a_{1}+a_{2}+a_{3}+\ldots+a_{q}}=\frac{p^{3}}{q^{3}} ; p \neq q\). Then \(\frac{a_{6}}{a_{21}}\) is equal to: [Online April 9, 2013] (a) \(\frac{41}{11}\) (b) \(\frac{31}{121}\) (c) \(\frac{11}{41}\) (d) \(\frac{121}{1861}\)

6 step solution

Problem 35

If 100 times the \(100^{\text {th }}\) term of an AP with non zero common difference equals the 50 times its \(50^{\text {th }}\) term, then the \(150^{\text {th }}\) term of this AP is : (a) \(-150\) (b) 150 times its \(50^{\text {th }}\) term (c) 150 (d) Zero

5 step solution

Problem 36

If the A.M. between \(p^{\text {th }}\) and \(q^{\text {th }}\) terms of an A.P. is equal to the A.M. between \(r^{\text {th }}\) and \(s^{\text {th }}\) terms of the same A.P., then \(p+q\) is equal to Online May 26, 2012] (a) \(r+s-1\) (b) \(r+s-2\) (c) \(r+s+1\) (d) \(r+s\)

7 step solution

Problem 37

Suppose \(\theta\) and \(\phi(\neq 0)\) are such that \(\sec (\theta+\phi), \sec \theta\) and \(\sec (\theta-\phi)\) are in A.P. If \(\cos \theta=k \cos \left(\frac{\phi}{2}\right)\) for some \(k\), then \(k\) is equal to [Online May 19, 2012] (a) \(\pm \sqrt{2}\) (b) \(\pm 1\) (c) \(\pm \frac{1}{\sqrt{2}}\) (d) \(\pm 2\)

7 step solution

Problem 38

Let \(a_{n}\) be the \(n^{\text {th }}\) term of an A.P. If \(\sum_{r=1}^{100} a_{2 r}=\alpha\) and \(\sum_{r=1}^{100} a_{2 r-1}=\beta\), then the common difference of the A.P. is [2011] (a) \(\alpha-\beta\) (b) \(\frac{\alpha-\beta}{100}\) (c) \(\beta-\alpha\) (d) \(\frac{\alpha-\beta}{200}\)

9 step solution

Problem 39

A man saves \(\ 200\) in each of the first three months of his service. In each of the subsequent months his saving increases by \(\mathbf{7} 40\) more than the saving of immediately previous month. His total saving from the start of service will be \(¥ 11040\) after (a) 19 months (b) 20 months (c) 21 months (d) 18 months

8 step solution

Problem 40

A person is to count 4500 currency notes. Let \(a_{n}\) denote the number of notes he counts in the \(n^{\text {th }}\) minute. If \(a_{1}=a_{2}\) \(=\ldots=a_{10}=150\) and \(a_{10}, a_{11}, \ldots\) are in an AP with common difference \(-2\), then the time taken by him to count all notes is \([\mathbf{2 0 1 0}]\) (a) 34 minutes (b) 125 minutes (c) 135 minutes (d) 24 minutes

8 step solution

Problem 41

Let \(a_{1}, a_{2}, a_{3} \ldots \ldots \ldots . .\) be terms on A.P. If \(\frac{a_{1}+a_{2}+\ldots \ldots \ldots a_{p}}{a_{1}+a_{2}+\ldots \ldots \ldots .+a_{q}}=\frac{p^{2}}{q^{2}}, p \neq q\), then \(\frac{a_{6}}{a_{21}}\) equals \([2006]\) (a) \(\frac{41}{11}\) (b) \(\frac{7}{2}\) (c) \(\frac{2}{7}\) (d) \(\frac{11}{41}\)

6 step solution

Problem 42

Let \(T_{\mathrm{r}}\) be the rth term of an A.P. whose first term is a and common difference is \(d\). If for some positive integers \(m, n, m \neq n, T_{m}=\frac{1}{n}\) and \(T_{n}=\frac{1}{m}\), then \(a-d\) equals [2004] (a) \(\frac{1}{m}+\frac{1}{n}\) (b) 1 (c) \(\frac{1}{m n}\) (d) 0

5 step solution

Problem 43

Let \(T_{\mathrm{r}}\) be the rth term of an A.P. whose first term is a and common difference is \(d\). If for some positive integers \(m, n, m \neq n, T_{m}=\frac{1}{n}\) and \(T_{n}=\frac{1}{m}\), then \(a-d\) equals [2004] (a) \(\frac{1}{m}+\frac{1}{n}\) (b) 1 (c) \(\frac{1}{m n}\) (d) 0

9 step solution

Problem 44

If \(1, \log _{9}\left(3^{1-x}+2\right), \log _{3}\left(4.3^{x}-1\right)\) are in A.P. then \(x\) equals [2002] (a) \(\log _{3} 4\) (b) \(1-\log _{3} 4\) (c) \(1-\log _{4} 3\) (d) \(\log _{4} 3\)

6 step solution

Problem 45

If \(f(x+y)=f(x) f(y)\) and \(\sum_{x=1}^{\infty} f(x)=2, x, y \in \mathrm{N}\), where \(\mathrm{N}\) is the set of all natural numbers, then the value of \(\frac{f(4)}{f(2)}\) is : [Sep. 06, 2020 (I)] (a) \(\frac{2}{3}\) (b) \(\frac{1}{9}\) (c) \(\frac{1}{3}\) (d) \(\frac{4}{9}\)

7 step solution

Problem 46

Let \(a, b, c, d\) and \(p\) be any non zero distinct real numbers such that \(\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c+c d) p+\left(b^{2}+c^{2}+\right.\) \(\left.d^{2}\right)=0 .\) Then : (a) \(a, c, p\) are in A.P. (b) \(a, c, p\) are in GP. (c) \(a, b, c, d\) are in G.P. (d) \(a, b, c, d\) are in A.P.

9 step solution

Problem 47

Suppose that a function \(f: \mathrm{R} \rightarrow\) R satisfies \(f(x+y)=f(x) f(y)\) for all \(x, y \in \mathrm{R}\) and \(f(\mathrm{a})=3\). If \(\sum_{i=1}^{\mathrm{n}} f(i)=363\), then \(\mathrm{n}\) is equal to \(\quad\) [NA Sep. 06, 2020 (II)]

7 step solution

Problem 48

If \(2^{10}+2^{9} \cdot 3^{1}+2^{8} \cdot 3^{2}+\ldots+2 \times 3^{9}+3^{10}=\mathrm{S}-2^{11}\) then \(\mathrm{S}\) is equal to: [Sep. \(\mathbf{0 5 , 2 0 2 0}\) (I)] (a) \(3^{11}-2^{12}\) (b) \(3^{11}\) (c) \(\frac{3^{11}}{2}+2^{10}\) (d) \(2 \cdot 3^{11}\)

5 step solution

Problem 49

If the sum of the second, third and fourth terms of a positive term G.P. is 3 and the sum of its sixth, seventh and eighth terms is 243 , then the sum of the first 50 terms of this G.P. is: [Sep. \(05, \mathbf{2 0 2 0}\) (II)] (a) \(\frac{1}{26}\left(3^{49}-1\right)\) (b) \(\frac{1}{26}\left(3^{50}-1\right)\) (c) \(\frac{2}{13}\left(3^{50}-1\right)\) (d) \(\frac{1}{13}\left(3^{50}-1\right)\)

8 step solution

Problem 50

Let \(\alpha\) and \(\beta\) be the roots of \(x^{2}-3 x+p=0\) and \(\gamma\) and \(\delta\) be the roots of \(x^{2}-6 x+q=0\). If \(\alpha, \beta, \gamma, \delta\) form a geometric progression. Then ratio \((2 q+p):(2 q-p)\) is: \([\) Sep. \(04,2020(\mathrm{I})]\) (a) \(3: 1\) (b) \(9: 7\) (c) \(5: 3\) (d) \(33: 31\)

10 step solution

Problem 51

The value of \((0.16) \log _{2.5}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots\right.\) to \(\left.\infty\right)\) is equal to [NA Sep. \(\mathbf{0 3}, \mathbf{2 0 2 0}\) (I)]

6 step solution

Problem 52

The sum of the first three terms of a G.P. is \(S\) and their product is \(27 .\) Then all such \(S\) lie in : \(\quad\) [Sep. 02, \(\mathbf{2 0 2 0}\) (I)] (a) \((-\infty,-9] \cup[3, \infty)\) (b) \([-3, \infty)\) (c) \((-\infty,-3] \cup[9, \infty)\) (d) \((-\infty, 9]\)

7 step solution

Problem 53

If \(|x|<1,|y|<1\) and \(x \neq y\), then the sum to infinity of the following series [Sep. 02, 2020 (I)] \((x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)+\ldots .\) is : (a) \(\frac{x+y-x y}{(1+x)(1+y)}\) (b) \(\frac{x+y+x y}{(1+x)(1+y)}\) (c) \(\frac{x+y-x y}{(1-x)(1-y)}\) (d) \(\frac{x+y+x y}{(1-x)(1-y)}\)

6 step solution

Show/ page