Chapter 3

Algebra for JEE ADVANCED · 114 exercises

Problem 1

For \(a, x>0\) prove that at the most one term of the G.P. \(\sqrt{a-x} \cdot \sqrt{x}, \sqrt{a+x}\) can be rational.

6 step solution

Problem 1

If \(\log _{2}\left(5 \times 2^{x}+1\right), \log _{4}\left(2^{1-x}+1\right)\) and 1 are in A.P., then \(x\) equals a. \(\log _{2} 5\) b. \(1-\log _{5} 2\) c. \(\log _{5} 2\) d. none of these

5 step solution

Problem 2

If the terms of the A.P. \(\sqrt{a-x}, \sqrt{x}, \sqrt{a+x}, \ldots\) are all inte- gers, where \(a, x>0\), then find the least composite value of \(a\).

7 step solution

Problem 2

If three positive real numbers \(a, b, c\) are in A.P. such that \(a b c=4\), then the minimum value of \(b\) is a. \(2^{1 / 3}\) b. \(2^{2 / 3}\) \(c \cdot 2^{1 / 2}\) d. \(2^{3 / 2}\)

5 step solution

Problem 2

If sum of an infinite G.P. \(p, 1,1 / p, 1 / p^{2}, \ldots\), is \(9 / 2\), then value of \(p\) is a. 2 b. \(3 / 2\) c. 3 d. \(9 / 2\)

8 step solution

Problem 3

Find a three-digit number such that its digits are in increasing G.P. (from left to right) and the digits of the number obtained from it by subtracting 100 form an A.P.

6 step solution

Problem 3

If \(\sum_{r=1}^{n} r(r+1)(2 r+3)=a n^{4}+b n^{3}+c n^{2}+d n+e\), then a. \(a-b=d-c\) b. \(e=0\) c. \(a, b-2 / 3, c-1\) are in A.P. d. \((b+d) / a\) is an integer

6 step solution

Problem 4

Along a road lies an odd number of stones placed at intervals of \(10 \mathrm{~m}\). These stones have to be assembled around the middle stone. A person can carry only one stone at a time. A man carried the job with one of the end stones by carrying them in succession. In carrying all the stones, he covered a distance of \(3 \mathrm{~km}\). Find the number of stones.

9 step solution

Problem 4

The largest term common to the sequences \(1,11,21,31, \ldots\) to 100 terms and \(31,36,41,46, \ldots\) to 100 terms is a. 381 b. 471 c. 281 d. none of these

6 step solution

Problem 4

The terms of an infinitely decreasing G.P. in which all the terms are positive, the first term is 4, and the difference between the third and fifth term is \(32 / 81\), then a. \(r=1 / 3\) b. \(r=2 \sqrt{2} / 3\) c. \(S_{\infty}=6\) d. none of these

7 step solution

Problem 5

If the sum of \(m\) terms of an A.P. is the same as the sum of its \(n\) terms, then the sum of its \((m+n)\) terms is a. \(m n\) b. \(-m n\) c. \(1 / m m\) d. 0

7 step solution

Problem 5

The consecutive digits of a three digit number are in G.P. If the middle digit be increased by 2, then they form an A.P. If 792 is subtracted from this, then we get the number constituting of same three digits but in reverse order. Then number is divisible by a. 7 b. 49 c. 19 d. none of these

9 step solution

Problem 6

Let \(x=1+3 a+6 a^{2}+10 a^{3}+\cdots,|a|<1\) \(y=1+4 b+10 b^{2}+20 b^{3}+\cdots,|b|<1\) Find \(S=1+3(a b)+5(a b)^{2}+\cdots\) in terms of \(x\) and \(y\).

7 step solution

Problem 6

If the sides of a right angled triangic are in A.P. then the sines of the acute angles are \(\begin{array}{ll}\text { a. } \frac{3}{5}, \frac{4}{5} & \text { b. } \frac{1}{\sqrt{3}}, \sqrt{\frac{2}{3}}\end{array}\) c. \(\frac{1}{2}, \frac{\sqrt{3}}{2}\) d. none of these

6 step solution

Problem 7

Given that \(x+y+z=15\) when \(a, x_{1} y, z, b\) are in A.P. and \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{5}{3}\) when \(a, x, y, z, b\) are in H.P. Then a. G.M. of \(a\) and \(b\) is 3 b. one possible value of \(a+2 b\) is 11 c. A.M. of \(a\) and \(b\) is 6 d. greatest value of \(a-b\) is 8

8 step solution

Problem 8

Find the sum \(\frac{3}{1 \times 2} \times \frac{1}{2}+\frac{4}{2 \times 3} \times\left(\frac{1}{2}\right)^{2}+\frac{5}{3 \times 4} \times\left(\frac{1}{2}\right)^{3}+\cdots\) to \(n\) terms.

6 step solution

Problem 8

150 workers were engaged to finish a piece of work in a certain number of days. Four workers stopped working on the second day, four more workers stopped their work on the third day and so on. It took 8 more days to finish the work. Then the number of days in which the work was completed is a. 29 days b. 24 days c. 25 days d. none of these

7 step solution

Problem 8

Let \(a_{1}, a_{2}, a_{3}, \ldots, a_{n}\) be in G.P. such that \(3 a_{1}+7 a_{2}+3 a_{3}-4 a_{5}\) \(=0\). Then common ratio of G.P. can be a. 2 b. \(\frac{3}{2}\) c. \(\frac{5}{2}\) d. \(-\frac{1}{2}\)

6 step solution

Problem 9

If \(S_{1}+S_{2}, S_{y}, \ldots, S_{n}\) are the sums of \(n\) terms of \(m\) A.P.'s whose first terms are \(1,2,3, \ldots, m\) and common differences are \(1,3,5, \ldots\) \((2 m-1)\), respectively, show that \(S_{1}+S_{2}+\cdots+S_{m}=\frac{m n}{2}(m n+1)\)

8 step solution

Problem 9

In an A.P. of which \(a\) is the first term, if the sum of the first \(p\) terms is zero, then the sum of the next \(q\) terms is a. \(-\frac{a(p+q) p}{q+1}\) b. \(\frac{a(p+q) p}{p+1}\) c. \(-\frac{a(p+q) q}{p-1}\) d. none of these

6 step solution

Problem 9

\(\frac{1}{\sqrt{2}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{8}}+\frac{1}{\sqrt{8}+\sqrt{11}}+\cdots n\) terms, is equal to a. \(\frac{\sqrt{3 n+2}-\sqrt{2}}{3}\) b. \(\frac{n}{\sqrt{2+3 n}+\sqrt{2}}\) c. less than \(n\) d. less than \(\sqrt{\frac{n}{2}}\)

5 step solution

Problem 10

If \(S_{[}, S_{2}\) and \(S_{3}\) be, respectively, the sum of \(n, 2 n\) and \(3 n\) terms of a G.P., prove that \(S_{1}\left(S_{3}-S_{2}\right)=\left(S_{2}-S_{1}\right)^{2}\).

5 step solution

Problem 10

If \(S_{u}\) denotes the sum of first ' \(n^{\prime}\) terms of an A.P. and \(\frac{S_{3_{n}}-S_{n-1}}{S_{2 n}-S_{2 n-1}}=31\), then the value of \(n\) is a. 21 b. 15 c. 16 d. 19

6 step solution

Problem 10

If \(a, b\), and \(c\) are in H.P. then the value of \(\frac{(a c+a b-b c)(a b+b c-a c)}{(a b c)^{2}}\) is a. \(\frac{(a+c)(3 a-c)}{4 a^{2} c^{2}}\) b. \(\frac{2}{b c}-\frac{1}{b^{2}}\) c. \(\frac{2}{b c}-\frac{1}{a^{2}}\) d. \(\frac{(a-c)(3 a+c)}{4 a^{2} c^{2}}\)

5 step solution

Problem 11

Find four numbers in a G.P. whose sum is 85 and product is \(4096 .\)

10 step solution

Problem 11

If \(a, b\), and \(c\) are in A.P. then \(a^{3}+c^{3}-8 b^{3}\) is equal to a. \(2 a b c\) b. \(6 a b c\) c. \(4 a b c\) d. none of these

7 step solution

Problem 12

There are \((4 n+1)\) terms in a certain sequence of which the first \((2 n+1)\) terms form an A.P. of common difference 2 and the last \((2 n+1)\) terms are in G.P. of common ratio \(1 / 2\). If the middle term of both A.P. and G.P. be the same, then find the mid-term of this sequence.

6 step solution

Problem 12

For an increasing A.P, \(a_{1}, a_{2}, \ldots, a_{n}\) if \(a_{1}+a_{3}+a_{5}=-12\) and \(a_{1} a_{3} a_{5}=80\), then which of the following is/are true? a. \(a_{1}=-10\) b. \(a_{2}=-1\) c. \(a_{3}=-4\) d. \(a_{5}=+2\)

5 step solution

Problem 13

If \(a, \frac{1}{b}, c\) and \(\frac{1}{p}, q, \frac{1}{r}\) form two arithmetic progressions of the same common difference, then \(a, q, c\) are in A.P. if a. \(p, b, r\) are in A.P. b. \(\frac{1}{p} ; \frac{1}{b}, \frac{1}{r}\) are in A.P. c. \(p, b, r\) are in G.P. d. none of these

6 step solution

Problem 14

Find the sum of \(n\) terms of the series whose \(n^{\text {th }}\) term is \(T(n)=\tan \frac{x}{2^{n}} \times \sec \frac{x}{2^{n-1}}\)

6 step solution

Problem 14

Suppose that \(F(n+1)=\frac{2 F(n)+1}{2}\) for \(n=1,2,3, \ldots\) and \(F(1)=2\). Then, \(F(101)\) equals a. 50 b. 52 c. 54 d. none of these

5 step solution

Problem 15

If the sum of \(n\) terms of an A.P. is \(c n(n-1)\), where \(c \neq 0\), then sum of the squares of these terms is a. \(c^{2} n(n+1)^{2}\) b. \(\frac{2}{3} c^{2} n(n-1)(2 n-1)\) c. \(\frac{2 c^{2}}{3} n(n+1)(2 n+1)\) d. none of these

6 step solution

Problem 16

Let \(a_{1}, a_{2}, \ldots, a_{n}\) be real numbers such that \(\sqrt{a_{1}}+\sqrt{a_{2}-1}+\sqrt{a_{3}-2}+\cdots+\sqrt{a_{n}-(n-1)}=\) \(\frac{1}{2}\left(a_{1}+a_{2}+\ldots+a_{n}\right)=\frac{n(n-3)}{4}\) Compute the value of \(\sum_{i=1}^{100} a_{i}\).

6 step solution

Problem 16

Consider an A.P. \(a_{1}, a_{2}, a_{3}, \ldots\) such that \(a_{3}+a_{5}+a_{\mathrm{a}}=11\) and \(a_{4}+a_{2}=-2\), then the value of \(a_{1}+a_{6}+a_{7}\) is a. \(-8\) b. 5 c. 7 d. 9

8 step solution

Problem 17

If the sum of \(n\) terms of an A.P. is given by \(S_{n}=a+b n+c n^{2}\), where \(a, b, c\) are independent of \(n\), then a. \(a=0\) b. common difference of A.P. must be \(2 b\) c. common difference of A.P. must be \(2 c\) d. first term of A.P. is \(b+c\)

4 step solution

Problem 18

Let \(a_{1}, a_{2}, a_{3}, \ldots\) be terms of an A.P. If \(\frac{a_{1}+a_{2}+\cdots+a_{p}}{a_{1}+a_{2}+\cdots+a_{9}}\) \(=\frac{p^{2}}{q^{2}}, p \neq q\), then \(\frac{a_{6}}{a_{21}}\) equals a. \(41 / 11\) b. \(7 / 2\) c. \(2 / 7\) d. \(11 / 41\)

7 step solution

Problem 18

If \(x^{2}+9 y^{2}+25 z^{2}=x y z\left(\frac{15}{x}+\frac{5}{y}+\frac{3}{z}\right)\), then a. \(x, y\), and \(z\) are in H.P. b. \(\frac{1}{x}, \frac{1}{y}, \frac{1}{z}\) are in A.P. c. \(x, y, z\) are in G.P. d. \(\frac{1}{x}, \frac{1}{y}, \frac{1}{z}\) are in G.P.

7 step solution

Problem 19

If \(a, b, c\), and \(d\) are four unequal positive numbers which are in A.P., then a. \(\frac{1}{a}+\frac{1}{d}>\frac{1}{b}+\frac{1}{c}\) b. \(\frac{1}{a}+\frac{1}{d}<\frac{1}{b}+\frac{1}{c}\) c. \(\frac{1}{b}+\frac{\mathrm{I}}{c}>\frac{4}{a+d}\) d. \(\frac{1}{a}+\frac{1}{d}=\frac{1}{b}+\frac{1}{c}\)

4 step solution

Problem 20

Three numbers form an increasing G.P. If the middle number is doubled, then the new numbers are in A.P. The common ratio of the G.P. is a. \(2-\sqrt{3} \quad\) b. \(2+\sqrt{3}\) c. \(\sqrt{3}-2\) d. \(3+\sqrt{2}\)

6 step solution

Problem 20

The next term of the G.P. \(x, x^{2}+2\), and \(x^{3}+10\) is a. \(\frac{729}{16}\) b. 6 c. 0 d. 54

5 step solution

Problem 21

If \(a_{1}, a_{2}, a_{3}\left(a_{1}>0\right)\) are three successive terms of a G.P. with common ratio \(r\), the value of \(r\) for which \(a_{3}>4 a_{2}-3 a_{1}\) holds is given by a. \(13\) or \(r<1\) d. none of these

7 step solution

Problem 23

If \(|a|<1\) and \(|b|<1\), then the sum of the series \(1+(1+a) b\) \(+\left(1+a+a^{2}\right) b^{2}+\left(1+a+a^{2}+a^{3}\right) b^{3}+\cdots\) is a. \(\frac{1}{(1-a)(1-b)}\) b. \(\frac{1}{(1-a)(1-a b)}\) c. \(\frac{1}{(1-b)(1-a b)}\) d. \(\frac{1}{(1-a)(1-b)(1-a b)}\)

6 step solution

Problem 23

Let \(S_{1}, S_{2}, \ldots .\) be squares such that for each \(n \geq 1\), the length of a side of \(S_{n}\) equals the length of a diagonal of \(S_{s+1^{*}}\) If the length of a side of \(S_{1}\) is \(10 \mathrm{~cm}\), then for which of the following values of \(n\) is the area of \(S_{n}\) less than 1 sq. \(\mathrm{cm} ?\) a. 7 b. 8 c. 9 d. 10

6 step solution

Problem 24

If \((p+q)^{\mathrm{th}}\) term of a G.P. is ' \(a\) ' and its \((p-q)^{\text {th }}\) term is ' \(b\) ' where \(a, b \in R^{+}\), then its \(p^{\text {it }}\) term is a. \(\sqrt{\frac{a^{3}}{b}}\) b. \(\sqrt{\frac{b^{3}}{a}}\) c. \(\sqrt{a b}\) d. none of these

7 step solution

Problem 24

If \(\frac{1}{b-a}+\frac{1}{b-c}=\frac{1}{a}+\frac{1}{c}\), then. a. \(a, b\), and \(c\) are in H.P. b. \(a, b\), and \(c\) are in A.P. c. \(b=a+c\) d. \(3 a=b+c\)

7 step solution

Problem 25

If the sides of a triangle are in G.P., and its largest angle is twice the smallest, then the common ratio \(r\) satisfies the inequality a. \(0

6 step solution

Problem 25

If \(a, b\), and \(c\) are in G.P. and \(x\) and \(y\), respectively, be arithmetic means between \(a, b\) and \(b, c\), then a. \(\frac{a}{x}+\frac{c}{y}=2\) b. \(\frac{a}{x}+\frac{c}{y}=\frac{c}{a}\) c. \(\frac{1}{x}+\frac{1}{y}=\frac{2}{b}\) d. \(\frac{1}{x}+\frac{1}{y}=\frac{2}{a c}\)

9 step solution

Problem 26

The value of \(0.2^{\log _{\sqrt{5}}\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots\right)}\), a. 4 b. \(\log 4\) c. \(\log 2\) d. none of these

5 step solution

Problem 26

Consider a sequence \(\left\\{a_{n}\right\\}\) with \(a_{1}=2\) and \(a_{n}=\frac{a_{n-1}^{2}}{a_{n-2}}\) for all \(n\) \(\geq 3\), terms of the sequence being distinct. Given that \(a_{2}\) and \(a_{5}\) are positive integers and \(a_{5} \leq 162\) then the possible value(s) of \(a_{5}\) can be a. 162 b. 64 c. 32 d. 2

7 step solution

Problem 27

If \((1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \cdots\left(1+x^{118}\right)=\sum_{r=0}^{n} x^{\prime}\), then \(n\) is equal a. 256 b. 255 c. 254 d. none of these

5 step solution

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