JEE Advanced Maths – Geometric Progression PYQ with Solutions

Q 1 :

In the quadratic equation $a x^{2} + b x + c = 0, Δ = b^{2} - 4 a c$ and $α + β, α^{2} + β^{2}, α^{3} + β^{3}$ are in G.P. where $α, β$ are the roots of $a x^{2} + b x + c = 0,$ then [2005]

$Δ \neq 0$
$b Δ = 0$
$c Δ = 0$
$Δ = 0$

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(3)

$In the quadratic equation a x^{2} + b x + c = 0$

$Δ = b^{2} - 4 a c and α + β = - \frac{b}{a}, α β = \frac{c}{a}$

$α^{2} + β^{2} = {(α + β)}^{2} - 2 α β$

$= \frac{b^{2}}{a^{2}} - \frac{2 c}{a} = \frac{b^{2} - 2 a c}{a^{2}}$

$and α^{3} + β^{3} = - \frac{b^{3}}{a^{3}} - \frac{3 c}{a} (- \frac{b}{a}) = - (\frac{b^{3} - 3 a b c}{a^{3}})$

$Since α + β, α^{2} + β^{2} and α^{3} + β^{3} are in G.P.$

$∴$ $- \frac{b}{a}, - \frac{b^{2} - 2 a c}{a^{2}}, - \frac{b^{3} - 3 a b c}{a^{3}} are in G.P.$

$\Rightarrow {(\frac{b^{2} - 2 a c}{a^{2}})}^{2} = \frac{b}{a} (\frac{b^{3} - 3 a b c}{a^{3}})$

$\Rightarrow b^{4} + 4 a^{2} c^{2} - 4 a b^{2} c = b^{4} - 3 a b^{2} c$

$\Rightarrow 4 a^{2} c^{2} - a b^{2} c = 0 \Rightarrow a c Δ = 0$

$\Rightarrow c Δ = 0$ $(∵ in quadratic equation a \neq 0)$

Q 2 :

An infinite G.P. has first term $' x'$ and sum '5', then $x$ belongs to [2004]

$x < - 10$
$- 10 < x < 0$
$0 < x < 10$
$x > 10$

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(3)

$\frac{x}{1 - r} = 5 \Rightarrow r = 1 - \frac{x}{5}$

$Since G.P. contains infinite terms$

$∴$ $- 1 < r < 1$

$\Rightarrow - 1 < 1 - \frac{x}{5} < 1 \Rightarrow - 2 < - \frac{x}{5} < 0$

$\Rightarrow - 10 < x < 0 \Rightarrow 0 < \frac{x}{5} < 2$

$∴$ $0 < x < 10$

Q 3 :

Suppose $a, b, c$ are in A.P. and $a^{2}, b^{2}, c^{2}$ are in G.P. If $a < b < c$ and $a + b + c = \frac{3}{2}$ , then the value of $a$ is [2002]

$\frac{1}{2 \sqrt{2}}$
$\frac{1}{2 \sqrt{3}}$
$\frac{1}{2} - \frac{1}{\sqrt{3}}$
$\frac{1}{2} - \frac{1}{\sqrt{2}}$

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(4)

$Since a, b, c are in A.P.$

$∴$ $2 b = a + c$

$But given a + b + c = \frac{3}{2} \Rightarrow b = \frac{1}{2} and then a + c = 1$

$Also a^{2}, b^{2}, c^{2} are in G.P. \Rightarrow b^{4} = a^{2} c^{2}$

$\Rightarrow b^{2} = \pm a c \Rightarrow a c = \frac{1}{4} or - \frac{1}{4}$

$and a + c = 1 . . . (i)$

$Considering a + c = 1 and a c = \frac{1}{4}$

${(a - c)}^{2} = 1 - 1 = 0 \Rightarrow a = c but a \neq c$

$as given that a < b < c$

$∴$ $a + c = 1 and a c = - \frac{1}{4}$

$\Rightarrow {(a - c)}^{2} = 1 + 1 = 2 \Rightarrow a - c = \pm \sqrt{2}$

$but a < c \Rightarrow a - c = - \sqrt{2} . . . (i i)$

$Solving (i) and (ii), we get a = \frac{1}{2} - \frac{1}{\sqrt{2}}$

Q 4 :

Let $α, β$ be the roots of $x^{2} - x + p = 0$ and $γ, δ$ be the roots of $x^{2} - 4 x + q = 0$ . If $α, β, γ, δ$ are in G.P., then the integral values of $p$ and $q$ respectively, are [2001]

$- 2, - 32$
$- 2, 3$
$- 6, 3$
$- 6, - 32$

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(1)

$α, β are the roots of x^{2} - x + p = 0$

$∴$ $α + β = 1 . . . (i)$

$α β = p . . . (i i)$

$γ, δ are the roots of x^{2} - 4 x + q = 0$

$∴$ $γ + δ = 4 . . . (i i i)$

$γ δ = q . . . (i v)$

$α, β, γ, δ are in G.P.$

$∴$ $Let α = a, β = a r, γ = a r^{2}, δ = a r^{3}$

$Substituting these values in equations (i), (ii), (iii) and (iv), we get$

$a + a r = 1 . . . (v)$
$a^{2} r = p . . . (v i)$
$a r^{2} + a r^{3} = 4 . . . (v i i)$
$a^{2} r^{5} = q . . . (v i i i)$

$On dividing (vii) by (v), we get$

$\frac{a r^{2} (1 + r)}{a (1 + r)} = \frac{4}{1} \Rightarrow r^{2} = 4 \Rightarrow r = 2, - 2$

$From (v), a = \frac{1}{1 + r} = \frac{1}{1 + 2} or \frac{1}{1 - 2} = \frac{1}{3} or - 1$

Since $p$ is an integer (given), $r$ is also an integer $(2 or - 2)$

$From (vi), a \neq \frac{1}{3} . Hence a = - 1 and r = - 2$

$∴$ $p = {(- 1)}^{2} \times (- 2) = - 2$

$q = {(- 1)}^{2} \times {(- 2)}^{5} = - 32$

Q 5 :

Consider an infinite geometric series with first term $a$ and common ratio $r$ . If its sum is 4 and the second term is $\frac{3}{4}$ , then [2000]

$a = \frac{4}{7}, r = \frac{3}{7}$
$a = 2, r = \frac{3}{8}$
$a = \frac{3}{2}, r = \frac{1}{2}$
$a = 3, r = \frac{1}{4}$

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(4)

$Sum = 4 and second term = \frac{3}{4}$

$\Rightarrow \frac{a}{1 - r} = 4 and a r = \frac{3}{4} \Rightarrow r = \frac{3}{4 a}$

$∴$ $\frac{a}{1 - \frac{3}{4 a}} = 4 \Rightarrow \frac{4 a^{2}}{4 a - 3} = 4$

$\Rightarrow a^{2} - 4 a + 3 = 0$ $∴$ $a = 1 or 3$

$When a = 1, r = \frac{3}{4} and when a = 3, r = \frac{1}{4}$

Q 6 :

Let $a_{1}, a_{2}, a_{3}, \dots$ be a sequence of positive integers in arithmetic progression with common difference 2. Also, let $b_{1}, b_{2}, b_{3}, \dots$ be a sequence of positive integers in geometric progression with common ratio 2. If $a_{1} = b_{1} = c,$ then the number of all possible values of c, for which the equality $2 (a_{1} + a_{2} + \dots + a_{n}) = b_{1} + b_{2} + \dots + b_{n}$ holds for some positive integer $n$ , is ______. [2020]

(1)

It is given that

$2 (a_{1} + a_{2} + \dots + a_{n}) = b_{1} + b_{2} + \dots + b_{n}$

$\Rightarrow 2 \times \frac{n}{2} (2 c + (n - 2) \cdot 2) = c (\frac{2^{n} - 1}{2 - 1})$ $[∵ a_{1} = c, b_{1} = c]$

$\Rightarrow c (2^{n} - 1 - 2 n) = 2 n^{2} - 2 n$

$\Rightarrow c = \frac{2 n^{2} - 2 n}{2^{n} - 1 - 2 n} \in ℕ$

$So, 2 n^{2} - 2 n \geq 2^{n} - 1 - 2 n$

$\Rightarrow 2 n^{2} + 1 \geq 2^{n} \Rightarrow n < 7$

$∵$ $c \in ℕ \Rightarrow c > 0 \Rightarrow n > 2$

$\Rightarrow n can be 3, 4, 5 or 6$

$Checking c against these values of n$

$When n = 3, c = 14$

$When n = 4, c = \frac{24}{7} which is not possible$

$When n = 5, c = \frac{40}{21} which is not possible$

$When n = 6, c = \frac{60}{51} which is not possible$

$∴$ $we get c = 12 when n = 3$

$Hence, there exists only one value of c which holds the inequality.$

Topic Question Set

Q 1 :

In the quadratic equation $a x^{2} + b x + c = 0, Δ = b^{2} - 4 a c$ and $α + β, α^{2} + β^{2}, α^{3} + β^{3}$ are in G.P. where $α, β$ are the roots of $a x^{2} + b x + c = 0,$ then [2005]

Q 2 :

An infinite G.P. has first term $' x'$ and sum '5', then $x$ belongs to [2004]

Q 3 :

Suppose $a, b, c$ are in A.P. and $a^{2}, b^{2}, c^{2}$ are in G.P. If $a < b < c$ and $a + b + c = \frac{3}{2}$ , then the value of $a$ is [2002]

Q 4 :

Let $α, β$ be the roots of $x^{2} - x + p = 0$ and $γ, δ$ be the roots of $x^{2} - 4 x + q = 0$ . If $α, β, γ, δ$ are in G.P., then the integral values of $p$ and $q$ respectively, are [2001]

Q 5 :

Consider an infinite geometric series with first term $a$ and common ratio $r$ . If its sum is 4 and the second term is $\frac{3}{4}$ , then [2000]

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Topic Question Set

Q 1 : In the quadratic equation ax2+bx+c=0, Δ=b2-4ac and α+β, α2+β2, α3+β3 are in G.P. where α,β are the roots of ax2+bx+c=0, then [2005]

Q 2 : An infinite G.P. has first term 'x' and sum '5', then x belongs to [2004]

Q 3 : Suppose a,b,c are in A.P. and a2,b2,c2 are in G.P. If a<b<c and a+b+c=32, then the value of a is [2002]

Q 4 : Let α,β be the roots of x2-x+p=0 and γ,δ be the roots of x2-4x+q=0. If α,β,γ,δ are in G.P., then the integral values of p and q respectively, are [2001]

Q 5 : Consider an infinite geometric series with first term a and common ratio r. If its sum is 4 and the second term is 34, then [2000]

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Q 1 :

In the quadratic equation $a x^{2} + b x + c = 0, Δ = b^{2} - 4 a c$ and $α + β, α^{2} + β^{2}, α^{3} + β^{3}$ are in G.P. where $α, β$ are the roots of $a x^{2} + b x + c = 0,$ then [2005]

Q 2 :

An infinite G.P. has first term $' x'$ and sum '5', then $x$ belongs to [2004]

Q 3 :

Suppose $a, b, c$ are in A.P. and $a^{2}, b^{2}, c^{2}$ are in G.P. If $a < b < c$ and $a + b + c = \frac{3}{2}$ , then the value of $a$ is [2002]

Q 4 :

Let $α, β$ be the roots of $x^{2} - x + p = 0$ and $γ, δ$ be the roots of $x^{2} - 4 x + q = 0$ . If $α, β, γ, δ$ are in G.P., then the integral values of $p$ and $q$ respectively, are [2001]

Q 5 :

Consider an infinite geometric series with first term $a$ and common ratio $r$ . If its sum is 4 and the second term is $\frac{3}{4}$ , then [2000]