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

Q 11 :
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 12 :
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]

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(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 13 :
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 14 :
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 15 :
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 16 :
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. of an increasing series]

Substituter 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 17 :
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)

$2 (a r^{2} - 9) = a - 7 + a r^{3} - 5$ ... (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 $\frac{1}{24} (x_{1} x_{2} x_{3} x_{4}) = \frac{(5184)}{24} = 216$ .

Q 18 :
If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eight, tenth and twelfth terms is 15309, then the sum of its first nine terms is : [2025]

760
755
750
757

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(4)

We have, $a r + a r^{3} + a r^{5} = 21, a r^{7} + a r^{9} + a^{11} = 15309$

$\Rightarrow a r (1 + r^{2} + r^{4}) = 21$ ... (i)

and $a r^{7} (1 + r^{2} + r^{4}) = 15309$ ... (ii)

Divide equation (ii) by (i), we get

$\frac{a \cdot r^{7}}{a r} = \frac{15309}{21} \Rightarrow r^{6} = 729$

$\Rightarrow r = 3 and a = \frac{7}{91}$

Now, $S_{9} = \frac{a \cdot (r^{9} - 1}{r - 1}$

$= \frac{\frac{7}{91} (19683 - 1)}{2}$

$= \frac{7 \times 19682}{91 \times 2}$

$= \frac{9841}{13} = 757$ .

Q 19 :
Let $a_{1}, a_{2}, a_{3}, . . . . .$ be a G.P. of increasing positive terms, If $a_{2} a_{5} = 28$ and $a_{2} a_{4} = 29$ , then $a_{6}$ is equal to : [2025]

812
526
784
628

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(3)

Here, r > 0 and $a_{1}, a_{2}, a_{3}, . . . . . > 0$ as G.P. has increasing positive terms.

$a_{1} a_{5} = 28$ (Given)

$\Rightarrow a (a r^{4}) = 28 = a^{2} a^{4} = 28$ ... (i)

Also, $a_{2} + a_{4} = 29$

$\Rightarrow a r + a r^{3} = 29$

$\Rightarrow a r (1 + r^{2}) = 29$ ... (i)

From (i) and (ii), we get

$\frac{r^{2}}{{(1 + r^{2})}^{2}} = \frac{28}{{(29)}^{2}}$

$\Rightarrow 841 r^{2} = 28 r^{4} + 56 r^{2} + 28$

$\Rightarrow 28 r^{4} - 785 r^{2} + 28 = 0$

$\Rightarrow (r^{2} - 28) (28 r^{2} - 1) = 0$

$\Rightarrow r^{2} = 28 or \frac{1}{28}$

$\Rightarrow r = \sqrt{28}$ $[∵ r \neq \frac{1}{28} as G . P . is increasing]$

$\Rightarrow a = \frac{1}{\sqrt{28}} and a_{6} = a r^{5} = \frac{1}{\sqrt{28}} \times {(\sqrt{28})}^{5} = 784$ .

Q 20 :
Let $< a_{n} >$ be a sequence such that $a_{0} = 0, a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}, n = 0, 1, 2, 3, . . .$ . Then $\sum_{k = 1}^{100} a_{k}$ is equal to . [2025]

$3 a_{99} - 100$
$3 a_{100} - 100$
$3 a_{100} + 100$
$3 a_{99} + 100$

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(2)

We have, $a_{0} = 0, a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}$

Let $2 x^{2} = 5 x - 3$

$\Rightarrow 2 x^{2} - 5 x + 3 = 0 \Rightarrow (2 x - 3) (x - 1) = 0$

$\Rightarrow x = 1, \frac{3}{2} ∴ a_{n} = A {(1)}^{n} + B {(\frac{3}{2})}^{n}$

Put n = 0; A + B

$n = 1; \frac{1}{2} = A + \frac{3}{2} B \Rightarrow B = 1, A = - 1$

$\Rightarrow a_{n} = (- 1) + {(\frac{3}{2})}^{n}$

$\Rightarrow \sum_{k = 1}^{100} a_{k} = \sum_{k = 1}^{100} (- 1) + {(\frac{3}{2})}^{k}$

$= - 100 + \frac{(\frac{3}{2}) ({(\frac{3}{2})}^{100} - 1)}{(\frac{3}{2} - 1)}$

$\Rightarrow - 100 + [3 ({(\frac{3}{2})}^{100} - 1)]$

$= 3 a_{100} - 100$ .

Topic Question Set

Q 11 :
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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Q 12 :
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]

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Q 13 :
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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Q 15 :
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 16 :
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 17 :
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]

Q 18 :
If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eight, tenth and twelfth terms is 15309, then the sum of its first nine terms is : [2025]

Q 19 :
Let $a_{1}, a_{2}, a_{3}, . . . . .$ be a G.P. of increasing positive terms, If $a_{2} a_{5} = 28$ and $a_{2} a_{4} = 29$ , then $a_{6}$ is equal to : [2025]

Q 20 :
Let $< a_{n} >$ be a sequence such that $a_{0} = 0, a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}, n = 0, 1, 2, 3, . . .$ . Then $\sum_{k = 1}^{100} a_{k}$ is equal to . [2025]

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Topic Question Set

Q 11 : If the range of f(θ)=sin4θ+3cos2θsin4θ+cos2θ, θ∈R is [α,β], then the sum of the infinite G.P., whose first term is 64 and the common ratio is αβ, is equal to _______. [2024]

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Q 12 : 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]

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Q 13 : If 8=3+14(3+p)+142(3+2p)+143(3+3p)+⋯∞, then the value of p is _______ . [2024]

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Q 14 : Let the coefficient of xr in the expansion of (x+3)n-1+(x+3)n-2(x+2)+(x+3)n-3(x+2)2+…+(x+2)n-1 be αr. If ∑r=0nαr=βn-γn, β,γ∈N, then the value of β2+γ2 equals ______ . [2024]

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Q 15 : 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 16 : Let a1,a2,a3,... be a G.P. of increasing positive numbers. If a3a5=729 and a2+a4=1114, then 24(a1+a2+a3) is equal to [2025]

Q 17 : Let x1,x2,x3,x4 be in a geometric progression. If 2, 7, 9, 5 are subtracted respectively from x1,x2,x3,x4, then the resulting numbers are in an arithmetic progression. Then the value of 124(x1x2x3x4) is : [2025]

Q 18 : If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eight, tenth and twelfth terms is 15309, then the sum of its first nine terms is : [2025]

Q 19 : Let a1,a2,a3, ..... be a G.P. of increasing positive terms, If a2a5=28 and a2a4=29, then a6 is equal to : [2025]

Q 20 : Let <an> be a sequence such that a0=0,a1=12 and 2an+2=5an+1–3an, n=0,1,2,3,.... Then ∑k=1100ak is equal to . [2025]

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Q 11 :
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]

Q 12 :
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]

Q 13 :
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]

Q 15 :
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 16 :
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 17 :
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]

Q 18 :
If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eight, tenth and twelfth terms is 15309, then the sum of its first nine terms is : [2025]

Q 19 :
Let $a_{1}, a_{2}, a_{3}, . . . . .$ be a G.P. of increasing positive terms, If $a_{2} a_{5} = 28$ and $a_{2} a_{4} = 29$ , then $a_{6}$ is equal to : [2025]

Q 20 :
Let $< a_{n} >$ be a sequence such that $a_{0} = 0, a_{1} = \frac{1}{2}$ and $2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}, n = 0, 1, 2, 3, . . .$ . Then $\sum_{k = 1}^{100} a_{k}$ is equal to . [2025]