Let a1- a2- a3- - be a G-P- such that a1-0- a1-a2-4 and a3-a4-16- If -9i-1ai-4 - then - is equal to -

Let the G.P. be a,ar,ar^2,... a_1+a_2=4 ⇒ a+ar=4 ⇒ a(1+r)=4 a_3+a_4=16 ⇒ ar^2+ar^3=16 ⇒ ar^2(1+r)=16 ⇒ 4r^2=16 ⇒ r=± 2 If r=2, a=43 which is not possible as a_ ...

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Byju's Answer

Standard XII

Mathematics

Sum of n Terms

Let a1, a2, a...

Question

Let $a_{1}, a_{2}, a_{3}, \dots$ be a G.P. such that $a_{1} < 0, a_{1} + a_{2} = 4$ and $a_{3} + a_{4} = 16$ . If $9 \sum i = 1 a_{i} = 4 λ$ , then $λ$ is equal to :

171

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\frac{511}{3}

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- 171

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- 513

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Solution

The correct option is C $- 171$
Let the G.P. be $a, a r, a r^{2}, . . .$
$a_{1} + a_{2} = 4 \Rightarrow a + a r = 4 \Rightarrow a (1 + r) = 4$
$a_{3} + a_{4} = 16$
$\Rightarrow a r^{2} + a r^{3} = 16 \Rightarrow a r^{2} (1 + r) = 16 \Rightarrow 4 r^{2} = 16$
$\Rightarrow r = \pm 2$

If $r = 2, a = \frac{4}{3}$ which is not possible as $a_{1} < 0$
If $r = - 2, a = - 4$
$9 \sum i = 1 a_{i} = \frac{a (r^{9} - 1)}{r - 1} = \frac{(- 4) [(- 2)^{9} - 1]}{- 3} = \frac{4}{3} (- 512 - 1) = 4 \times (- 171)$

$∴ λ = - 171$

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Let a1,a2,a3,… be a G.P. such that a1<0, a1+a2=4 and a3+a4=16. If 9∑i=1ai=4λ, then λ is equal to :

The correct option is C −171Let the G.P. be a,ar,ar2,... a1+a2=4⇒a+ar=4⇒a(1+r)=4 a3+a4=16 ⇒ar2+ar3=16⇒ar2(1+r)=16⇒4r2=16 ⇒r=±2 If r=2, a=43 which is not possible as a1<0 If r=−2, a=−4 9∑i=1ai=a(r9−1)r−1 =(−4)[(−2)9−1]−3 =43(−512−1)=4×(−171) ∴λ=−171

Let $a_{1}, a_{2}, a_{3}, \dots$ be a G.P. such that $a_{1} < 0, a_{1} + a_{2} = 4$ and $a_{3} + a_{4} = 16$ . If $9 \sum i = 1 a_{i} = 4 λ$ , then $λ$ is equal to :