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Previous Year Questions

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Previous Year Questions
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Phrases Previous Year Questions (PYQs)

Phrases Progressions PYQ

Qus : 1

The sum of infinite terms of a decreasing GP is equal to the greatest value of the function

in the interval [-2,3] and the difference between the first two terms is f'(0). Then the common ratio of GP is

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Solution

Qus : 2

$a, b, c$ are in GP and

$log a - log 2b$ ,

$log 2b - log 3c$ and

$log 3c - log a$ are in AP, then

$a, b, c$ are the lengths of the sides of a triangle which is

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Solution

Qus : 3

Not Available

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Solution

Qus : 4

$a, a, a_2, ., a_{2n-1},b$ are in AP,

$a, b_1, b_2,...b_{2n-1}, b$ are in GP and

$a, c_1, c_2,... c_{2n-1}, b$ are in HP, where a, b are positive, then the equation

$a_n x^2-b_n+c_n$ has its roots

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Solution

Qus : 5

$x=1+\sqrt[{6}]{2}+\sqrt[{6}]{4}+\sqrt[{6}]{8}+\sqrt[{6}]{16}+\sqrt[{6}]{32}$ then

${\Bigg{(}1+\frac{1}{x}\Bigg{)}}^{24}$ =

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Solution

Given:

$x = 1 + 2^{1/6} + 4^{1/6} + 8^{1/6} + 16^{1/6} + 32^{1/6}$

Step 1: Write in powers of $a = 2^{1/6}$

$x = 1 + a + a^2 + a^3 + a^4 + a^5 = 1 + \frac{a(a^5 - 1)}{a - 1}$

Step 2: Use identity $a^6 = 2 \Rightarrow a^5 = \frac{2}{a}$

$x = 1 + \frac{2 - a}{a - 1} = \frac{1}{a - 1} \Rightarrow 1 + \frac{1}{x} = a \Rightarrow \left(1 + \frac{1}{x} \right)^{24} = a^{24}$

Step 3: Final calculation

$a = 2^{1/6} \Rightarrow a^{24} = (2^{1/6})^{24} = 2^4 = \boxed{16}$

✅ Final Answer: $\boxed{16}$

Qus : 6

The number of solutions of

${5}^{1+|\sin x|+|\sin x{|}^2+\ldots}=25$ for

$x\in(-\mathrm{\pi},\mathrm{\pi})$ is

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Solution

Step 1: Recognize the series

This is a geometric series with first term $a = 1$ , common ratio $r = |\sin x| \in [0,1]$ , so: $\text{Sum} = \frac{1}{1 - |\sin x|}$

Step 2: Rewrite the equation

$5^{\frac{1}{1 - |\sin x|}} = 25 = 5^2$

Equating exponents: $\frac{1}{1 - |\sin x|} = 2 \Rightarrow 1 - |\sin x| = \frac{1}{2} \Rightarrow |\sin x| = \frac{1}{2}$

Step 3: Solve for $x \in (-\pi, \pi)$

We want all $x \in (-\pi, \pi)$ such that $|\sin x| = \frac{1}{2}$

So $\sin x = \pm \frac{1}{2}$ . Within $(-\pi, \pi)$ , the values of $x$ satisfying this are:

$x = \frac{\pi}{6}$
$x = \frac{5\pi}{6}$
$x = -\frac{\pi}{6}$
$x = -\frac{5\pi}{6}$

✅ Final Answer: $\boxed{4}$ solutions

Qus : 7

Which term of the series

$\frac{\sqrt[]{5}}{3},\, \frac{\sqrt[]{5}}{4},\frac{1}{\sqrt[]{5}},\, ...$ is

$\frac{\sqrt{5}}{13}$ ?

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Solution

Qus : 8

In a Harmonic Progression,

$p^{th}$ term is

$q$ and the

$q^{th}$ term is

$p$ . Then

$pq^{th}$ term is

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Solution

Qus : 9

If one AM (Arithmetic mean) 'a' and two GM's (Geometric means) p and q be inserted between any two positive numbers, the value of p^3+q^3 is

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Solution

Problem:

If one Arithmetic Mean (AM) $a$ and two Geometric Means $p$ and $q$ are inserted between any two positive numbers, find the value of: $p^3 + q^3$

Given:

Let two positive numbers be $A$ and $B$ .
One AM: $a = \frac{A + B}{2}$
Two GMs inserted: so the four terms in G.P. are: $A, \ p = \sqrt[3]{A^2B}, \ q = \sqrt[3]{AB^2}, \ B$

Now calculate:

$pq = \sqrt[3]{A^2B} \cdot \sqrt[3]{AB^2} = \sqrt[3]{A^3B^3} = AB$
$p^3 = A^2B, \quad q^3 = AB^2$
$p^3 + q^3 = A^2B + AB^2 = AB(A + B)$

Also,

$2apq = 2 \cdot \frac{A + B}{2} \cdot AB = AB(A + B)$

✅ Therefore,

$\boxed{p^3 + q^3 = 2apq}$

Qus : 10

The number of common terms in the two sequences 17, 21, 25, ..........., 817 and 16, 21, 26, ..........., 851 is

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Solution

Qus : 11

Sum to infinity of a geometric is twice the sum of the first two terms. Then what are the possible values of common ratio?

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Solution

Qus : 12

Suppose that m and n are fixed numbers such that the mth term am is equal to n and nth term an is equal to m, (m≠n), the the value of (m+n)th term is

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Solution

Qus : 13

The harmonic mean of two numbers is 4. Their arithmetic mean A and the geometric mean G satisfy the relation 2A+G² = 27, then the two numbers are

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Solution

Qus : 14

Three positive number whose sum is 21 are in arithmetic progression. If 2, 2, 14 are added to them respectively then resulting numbers are in geometric progression. Then which of the following is not among the three numbers?

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Solution

Let the three terms in A.P. be a – d, a, a + d.

given that a – d + a + a + d = 21

a = 7

then the three term in A.P. are 7 – d, 7, 7 + d

According to given condition 9 – d, 9, 21 + d are in G.P.

(9)² = (9 – d) (21 + d)

81 = 189 + 9d – 21d – d²

81 = 189 – 12d – d²

d² + 12d – 108 = 0

d(d + 18) – 6 (d + 18) = 0

(d – 6) (d + 18) = 0

We get, d = 6, –18

Putting d = 6 in the term 7 – d, 7, 7 + d we get 1, 7, 13.

Qus : 15

$H_1,H_2,\ldots,H_n$ are n harmonic means between a and b

$(b\ne a)$ ;,then

$\frac{{{H}}_n+a}{{{H}}_n-a}+\frac{{{H}}_n+b}{{{H}}_n-b}$

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Solution

Qus : 16

are positive real numbers whose product is a fixed number c, then the minimum of

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Solution

Qus : 17

The four geometric means between 2 and 64 are

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Solution

Qus : 18

If a, b, c are in geometric progression, then

$log_{ax}^{a}, log_{bx}^{a}$ and

$log_{cx}^{a}$ are in

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Solution

Qus : 19

The value of the sum

$\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{25\sqrt{24}+24\sqrt{25}}$ is

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Solution

Qus : 20

An arithmetic progression has 3 as its first term. Also, the sum of the first 8 terms is twice the sum of the first 5 terms. Then what is the common difference?

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Solution

Qus : 21

The sum of infinite terms of decreasing GP is equal to the greatest value of the function

$f(x) = x^3 + 3x – 9$ in the interval [–2, 3] and difference between the first two terms is f '(0). Then the common ratio of the GP is

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Solution

? GP and Function Relation

Given: $f(x) = x^3 + 3x - 9$

The sum of infinite GP = max value of $f(x)$ on [−2, 3]

The difference between first two terms = $f'(0)$

Step 1: $f(x)$ is increasing ⇒ Max at $x = 3$

$f(3) = 27 \Rightarrow \frac{a}{1 - r} = 27$

Step 2: $f'(x) = 3x^2 + 3 \Rightarrow f'(0) = 3$

⇒ $a(1 - r) = 3$

Step 3: Solve:
$a = 27(1 - r)$
$\Rightarrow 27(1 - r)^2 = 3 \Rightarrow (1 - r)^2 = \frac{1}{9} \Rightarrow r = \frac{2}{3}$

✅ Final Answer: $r = \frac{2}{3}$

Qus : 22

If a, b, c, d are in HP and arithmetic mean of ab, bc, cd is 9 then which of the following number is the value of ad?

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Solution

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Phrases Previous Year Questions (PYQs)

Phrases Progressions PYQ

Solution

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Solution

Solution

Solution

Given:

Step 1: Write in powers of a=21/6 a = 2^{1/6}

Step 2: Use identity a6=2⇒a5=2a a^6 = 2 \Rightarrow a^5 = \frac{2}{a}

Step 3: Final calculation

Solution

Step 1: Recognize the series

Step 2: Rewrite the equation

Step 3: Solve for x∈(−π,π) x \in (-\pi, \pi)

Solution

Solution

Solution

Problem:

Given:

Now calculate:

Also,

✅ Therefore,

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Solution

? GP and Function Relation

Solution

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Edit

Step 1: Write in powers of $a = 2^{1/6}$

Step 2: Use identity $a^6 = 2 \Rightarrow a^5 = \frac{2}{a}$

Step 3: Solve for $x \in (-\pi, \pi)$