Sequences and Series jee mains questions | Sequences and Series Previous Year Question

Q 21 :

Let $a$ and $b$ be two distinct positive real numbers. Let the 11th term of a GP, whose first term is $a$ and third term is $b,$ is equal to $p^{t h}$ term of another GP, whose first term is $a$ and fifth term is $b .$ Then $p$ is equal to [2024]

21
24
20
25

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(1)

Given, $T_{1} = a, T_{3} = a r_{1}^{2} = b$

$\Rightarrow r_{1} = {(\frac{b}{a})}^{1 / 2}$ ...(i)

Also, $T_{1} = a, T_{5} = b$

$\Rightarrow a r_{2}^{4} = b \Rightarrow r_{2} = {(\frac{b}{a})}^{1 / 4}$ ...(ii)

And $11^{t h}$ term of first GP = $p^{t h}$ term of second GP

Now, $a r_{1}^{10} = a r_{2}^{p - 1}$

$\Rightarrow a {(\frac{b}{a})}^{5} =$ $a$ ${{(\frac{b}{a})}^{1 / 4}}^{p - 1}$ (Using (i) and (ii))

$\Rightarrow {(\frac{b}{a})}^{5} = {(\frac{b}{a})}^{\frac{p - 1}{4}} \Rightarrow 5 = \frac{p - 1}{4} \Rightarrow p - 1 = 20 \Rightarrow p = 21$

Q 22 :

For $0 < c < b < a,$ let $(a + b - 2 c) x^{2} + (b + c - 2 a) x + (c + a - 2 b) = 0$ and $α \neq 1$ be one of its root. Then, among the two statements [2024]

(I) If $α \in (- 1, 0),$ then $b$ cannot be the geometric mean of $a$ and $c$

(II) If $α \in (0, 1),$ then $b$ may be the geometric mean of $a$ and $c$

only (II) is true
Both (I) and (II) are true
only (I) is true
Neither (I) nor (II) is true

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(2)

We have, $(a + b - 2 c) x^{2} + (b + c - 2 a) x + (c + a - 2 b) = 0$

Put $x = 1$

$a + b - 2 c + b + c - 2 a + c + a - 2 b = 0$

$0 = 0$

$∴$ $x = 1$ is another root

$∴$ $α \cdot 1 = \frac{c + a - 2 b}{a + b - 2 c} \Rightarrow α = \frac{c + a - 2 b}{a + b - 2 c}$

If $- 1 < α < 0,$ then $- 1 < \frac{c + a - 2 b}{a + b - 2 c} < 0$

$\Rightarrow 0 < \frac{c + a - 2 b + a + b - 2 c}{a + b - 2 c} < 1 \Rightarrow 0 < \frac{2 a - b - c}{a + b - 2 c} < 1$

Since, $a > b > c > 0 \Rightarrow a + b > c + c$

$\Rightarrow a + b > 2 c \Rightarrow a + b - 2 c > 0$

$∴$ $2 a - b - c > 0 \Rightarrow 2 a > b + c \Rightarrow a > \frac{b + c}{2}$

$∴$ $b$ can not be the geometric mean of $a$ and $c$

If $0 < α < 1,$ then $0 < \frac{c + a - 2 b}{a + b - 2 c} < 1$

$\Rightarrow 0 < c + a - 2 b and c + a - 2 b < a + b - 2 c$

$\Rightarrow c < b$

$2 b < c + a, b < \frac{c + a}{2}$

$∴$ $b$ may be the geometric mean of $a$ and $c .$

Q 23 :

Let $2^{nd}$ , $8^{th}$ and $44^{th}$ terms of a non-constant A.P. be respectively the $1^{st}$ , $2^{nd}$ and $3^{rd}$ terms of a G.P. If the first term of the A.P. is 1, then the sum of its first 20 terms is equal to [2024]

990
980
970
960

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(3)

We have, first term of an A.P. =1.

Let the $1^{st}$ , $2^{nd}$ and $3^{rd}$ terms of a G.P. are $\frac{a}{r}, a, a r$ respectively.

Now, $T_{2} = \frac{a}{r} \Rightarrow 1 + (2 - 1) d = \frac{a}{r} \Rightarrow 1 + d = \frac{a}{r}$ ...(i)

$T_{8} = a \Rightarrow 1 + 7 d = a$ ...(ii)

$T_{44} = a r \Rightarrow 1 + 43 d = a r$ ...(iii)

Using equations (i) and (ii), we get

$\frac{1 + d}{1 + 7 d} = \frac{1}{r}$ ...(iv)

From (ii) and (iii), we get

$\frac{1 + 7 d}{1 + 43 d} = \frac{1}{r}$ ...(v)

$∴$ $\frac{1 + d}{1 + 7 d} = \frac{1 + 7 d}{1 + 43 d}$

$\Rightarrow 1 + 43 d + d + 43 d^{2} = 1 + 14 d + 49 d^{2}$

$\Rightarrow 6 d^{2} - 30 d = 0 \Rightarrow 6 d (d - 5) = 0 \Rightarrow d = 0, d = 5$

$∴ d = 5 (∵ d \neq 0)$

$∴ S_{20} = \frac{20}{2} [2 + 19 \times 5] = 10 [2 + 95] = 970$

Q 24 :

If the range of $f (θ) = \frac{\sin^{4} θ + 3 \cos^{2} θ}{\sin^{4} θ + \cos^{2} θ}, θ \in R$ is $[α, β],$ then the sum of the infinite G.P., whose first term is 64 and the common ratio is $\frac{α}{β},$ is equal to _______. [2024]

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(96)

We have, $f (θ) = \frac{\sin^{4} θ + 3 \cos^{2} θ}{\sin^{4} θ + \cos^{2} θ}$

$= 1 + \frac{2 \cos^{2} θ}{\sin^{4} θ + \cos^{2} θ} = 1 + \frac{2 \cos^{2} θ}{1 + \cos^{4} θ - \cos^{2} θ}$

$= 1 + \frac{2}{\frac{1}{\cos^{2} θ} + \cos^{2} θ - 1}$

Now, $\cos^{2} θ + \frac{1}{\cos^{2} θ} \geq 2$ $[∵ A . M . \geq G . M .]$

$\Rightarrow \frac{1}{\cos^{2} θ} + \cos^{2} θ - 1 \geq 1 \Rightarrow \cos^{2} θ + \frac{1}{\cos^{2} θ} - 1 \in [1, \infty)$

$\Rightarrow \frac{1}{\cos^{2} θ + \frac{1}{\cos^{2} θ} - 1} \in (0, 1]$

When $\cos θ = 0, f (θ) = 1$

$∴$ $f (θ) \in [1, 3] \Rightarrow α = 1, β = 3$

Sum of infinite G.P. with first term 64 and common ratio

$\frac{1}{3} = \frac{64}{1 - \frac{1}{3}} = 32 \times 3 = 96$

Q 25 :

If three successive terms of a G.P. with common ratio $r (r > 1)$ are the lengths of the sides of a triangle and $[r]$ denotes the greatest integer less than or equal to $r,$ then $3 [r] + [- r]$ is equal to ______. [2024]

0%

(1)

Let $a, a r$ and $a r^{2}$ be the three sides of the triangle.

Now, $a + a r > a r^{2}$

$\Rightarrow r^{2} - r - 1 < 0 \Rightarrow r \in (\frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2})$

So, $3 [r] + [- r] = 3 + (- 2) = 1$

Q 26 :

If $8 = 3 + \frac{1}{4} (3 + p) + \frac{1}{4^{2}} (3 + 2 p) + \frac{1}{4^{3}} (3 + 3 p) + \dots \infty,$ then the value of $p$ is _______ . [2024]

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(9)

$8 = 3 + \frac{1}{(4)} (3 + p) + \frac{1}{{(4)}^{2}} (3 + 2 p) + \frac{1}{{(4)}^{3}} (3 + 3 p) + \dots \infty$ ...(i)

Multiplying both sides by $\frac{1}{4},$ we get $2 = \frac{3}{4} + \frac{3 + p}{{(4)}^{2}} + \frac{3 + 2 p}{{(4)}^{3}} + \dots + \infty$ ...(ii)

Subtracting (ii) from (i), we get $6 = 3 + \frac{p}{4} + \frac{p}{4^{2}} + \dots$

$\Rightarrow 3 = p [\frac{1}{4} + \frac{1}{{(4)}^{2}} + \frac{1}{{(4)}^{3}} + \dots + \infty]$

$\Rightarrow 3 = p [\frac{\frac{1}{4}}{1 - \frac{1}{4}}]$ $[∵ S_{\infty} = \frac{a}{1 - r}$ for infinite geometric series $]$

$\Rightarrow 3 = p [\frac{1}{4} \times \frac{4}{3}] \Rightarrow p = 9$

Q 27 :

Let the coefficient of $x^{r}$ in the expansion of ${(x + 3)}^{n - 1} + {(x + 3)}^{n - 2} (x + 2) + {(x + 3)}^{n - 3} {(x + 2)}^{2} + \dots + {(x + 2)}^{n - 1}$ be $α_{r} .$ If $\sum_{r = 0}^{n} α_{r} = β^{n} - γ^{n}, β, γ \in N,$ then the value of $β^{2} + γ^{2}$ equals ______ . [2024]

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(25)

We have,

${(x + 3)}^{n - 1} + {(x + 3)}^{n - 2} (x + 2) + {(x + 3)}^{n - 3} {(x + 2)}^{2} + \dots + {(x + 2)}^{n - 1}$

$∴$ $\sum_{r = 0}^{n} α_{r} = 4^{n - 1} + 4^{n - 2} \times 3 + 4^{n - 3} \times 3^{2} + \dots + 3^{n - 1}$

$= 4^{n - 1} [1 + \frac{3}{4} + {(\frac{3}{4})}^{2} + \dots + {(\frac{3}{4})}^{n - 1}]$

$= 4^{n - 1} \times \frac{1 - {(\frac{3}{4})}^{n}}{1 - \frac{3}{4}} = 4^{n - 1} (1 - {(\frac{3}{4})}^{n}) (4)$

$= 4^{n} - 3^{n} = β - γ \Rightarrow b = 4, γ = 3$

$∴$ $β^{2} + γ^{2} = 16 + 9 = 25$

Q 28 :

Let ABC be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle ABC and the same process is repeated infinitely many times. If P is the sum of perimeters and Q is the sum of areas of all the triangles formed in this process, then: [2024]

$P^{2} = 6 \sqrt{3} Q$
$P^{2} = 72 \sqrt{3} Q$
$P^{2} = 36 \sqrt{3} Q$
$P = 36 \sqrt{3} Q^{2}$

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(3)

$∆ A B C is an equilateral triangle of side' a' unit.$

$Perimeter of ∆ A B C = 3 a$

$Area of ∆ A B C = \frac{\sqrt{3}}{4} a^{2}$

Now, $∆ D E F$ is made by joining midpoints of the sides of $∆ A B C$

$∴$ $D E = E F = D F = \frac{a}{2}$

$Perimeter of ∆ D E F = \frac{3 a}{2}$

$Area of ∆ D E F = \frac{\sqrt{3}}{4} \times \frac{a^{2}}{4} = \frac{\sqrt{3}}{16} a^{2}$

$∴$ $P = 3 a + \frac{3 a}{2} + \frac{3 a}{4} + \dots$

$= 3 a [1 + \frac{1}{2} + \frac{1}{2^{2}} + \dots] = \frac{3 a}{1 - \frac{1}{2}} = 6 a$

Now, $Q = \frac{\sqrt{3}}{4} a^{2} [1 + \frac{1}{4} + \frac{1}{4^{2}} + \dots] = \frac{\frac{\sqrt{3}}{4} a^{2}}{1 - \frac{1}{4}} = \frac{4}{3} \times \frac{\sqrt{3}}{4} a^{2}$

$= \frac{1}{\sqrt{3}} a^{2} = \frac{1}{\sqrt{3}} {(\frac{P}{6})}^{2}$ $[∵ P = 6 a]$

$\Rightarrow 36 \sqrt{3} Q = P^{2}$

Q 29 :

Let $a_{1}, a_{2}, a_{3}, . . .$ be a G.P. of increasing positive numbers. If $a_{3} a_{5} = 729$ and $a_{2} + a_{4} = \frac{111}{4}$ , then $24 (a_{1} + a_{2} + a_{3})$ is equal to [2025]

131
129
128
130

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(2)

Let $a$ be the first term and $r$ be the common ratio of GP respectively.

Given, $a_{3} a_{5} = 729 \Rightarrow a r^{2} \cdot a r^{4} = 729$

$\Rightarrow a^{2} r^{6} = 729 \Rightarrow a r^{3} = 27$ ... (i)

and $a_{2} + a_{4} = a r + a r^{3} = \frac{111}{4}$

$\Rightarrow a r + 27 = \frac{111}{4} \Rightarrow a r = \frac{3}{4}$ ... (ii)

Now, divide equation (i) by equation (ii), we get

$\frac{a r^{3}}{a r} = \frac{27}{3 / 4}$

$\Rightarrow r^{2} = 36 \Rightarrow r = 6$ [ $∵$ G.P. is an increasing series]

Substitute $r$ = 6 in equation (ii), we get

$a = \frac{1}{8}$

Now, $24 (a_{1} + a_{2} + a_{3})$

$= 24 (a + a r + a r^{2})$

$= 24 a (1 + r + r^{2})$

$= 24 \times \frac{1}{8} (1 + 6 + 36)$

= 3(43) = 129.

Q 30 :

Let $x_{1}, x_{2}, x_{3}, x_{4}$ be in a geometric progression. If 2, 7, 9, 5 are subtracted respectively from $x_{1}, x_{2}, x_{3}, x_{4}$ , then the resulting numbers are in an arithmetic progression. Then the value of $\frac{1}{24} (x_{1} x_{2} x_{3} x_{4})$ is : [2025]

36
216
72
18

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(2)

Given, $x_{1}, x_{2}, x_{3}, x_{4}$ be in a G.P.

$x_{1} = a, x_{2} = a r, x_{3} = a r^{2}, x_{4} = a r^{3}$

According to question

$a - 2, a r - 7, a r^{2} - 9, a r^{3} - 5$ are in A.P.

So, $2 (a r - 7) = a - 2 + a r^{2} - 9$ ... (i)

$2 (a r^{2} - 9) = a r - 7 + a r^{3} - 5$ ... (ii)

Solving (i) and (ii), we get r = 2, a = – 3

Product $= x_{1} x_{2} x_{3} x_{4} = a^{4} r^{6} = 81 \times 64 = 5184$

$∴$ The value of $\frac{1}{24} (x_{1} x_{2} x_{3} x_{4}) = \frac{(5184)}{24} = 216$ .

Topic Question Set

Q 21 :

Let $a$ and $b$ be two distinct positive real numbers. Let the 11th term of a GP, whose first term is $a$ and third term is $b,$ is equal to $p^{t h}$ term of another GP, whose first term is $a$ and fifth term is $b .$ Then $p$ is equal to [2024]

Q 23 :

Let $2^{nd}$ , $8^{th}$ and $44^{th}$ terms of a non-constant A.P. be respectively the $1^{st}$ , $2^{nd}$ and $3^{rd}$ terms of a G.P. If the first term of the A.P. is 1, then the sum of its first 20 terms is equal to [2024]

Q 24 :

If the range of $f (θ) = \frac{\sin^{4} θ + 3 \cos^{2} θ}{\sin^{4} θ + \cos^{2} θ}, θ \in R$ is $[α, β],$ then the sum of the infinite G.P., whose first term is 64 and the common ratio is $\frac{α}{β},$ is equal to _______. [2024]

0%

Q 25 :

If three successive terms of a G.P. with common ratio $r (r > 1)$ are the lengths of the sides of a triangle and $[r]$ denotes the greatest integer less than or equal to $r,$ then $3 [r] + [- r]$ is equal to ______. [2024]

0%

Q 26 :

If $8 = 3 + \frac{1}{4} (3 + p) + \frac{1}{4^{2}} (3 + 2 p) + \frac{1}{4^{3}} (3 + 3 p) + \dots \infty,$ then the value of $p$ is _______ . [2024]

0%

0%

Q 28 :

Let ABC be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle ABC and the same process is repeated infinitely many times. If P is the sum of perimeters and Q is the sum of areas of all the triangles formed in this process, then: [2024]

Q 29 :

Let $a_{1}, a_{2}, a_{3}, . . .$ be a G.P. of increasing positive numbers. If $a_{3} a_{5} = 729$ and $a_{2} + a_{4} = \frac{111}{4}$ , then $24 (a_{1} + a_{2} + a_{3})$ is equal to [2025]

Q 30 :

Let $x_{1}, x_{2}, x_{3}, x_{4}$ be in a geometric progression. If 2, 7, 9, 5 are subtracted respectively from $x_{1}, x_{2}, x_{3}, x_{4}$ , then the resulting numbers are in an arithmetic progression. Then the value of $\frac{1}{24} (x_{1} x_{2} x_{3} x_{4})$ is : [2025]

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