Let a1, a2, a3, ... be a G. P. such that a1

Question

Let a1, a2, a3, ... be a G. P. such that a1 < 0, a1 +a2 = 4 and a3 + a4 = 16. If displaystyle ∑i=19ai=4λ, then λ is equal to : (A) 171 (B) (511/3) (C) -171 (D) 513

Tardigrade Admin · Accepted Answer

a1+a2=4 r2a1+r2a2=16 ⇒ r2=4 ⇒ r=-2 as a1 < 0 and a1+a2=4 a1+a2(-2)=4 ⇒ a1=-4 4λ=(-4)(((-2)9-1/-2-1))=(-4)×(513/3) ⇒ λ=-171

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

$171$

$\frac{511}{3}$

$- 171$

$513$

$a_{1} + a_{2} = 4$
$r^{2} a_{1} + r^{2} a_{2} = 16$
$\Rightarrow r^{2} = 4 \Rightarrow r = - 2 a s a_{1} < 0$
and $a_{1} + a_{2} = 4$
$a_{1} + a_{2} (- 2) = 4 \Rightarrow a_{1} = - 4$
$4 λ = (- 4) (\frac{( - 2 ) ^{9} - 1}{- 2 - 1}) = (- 4) \times \frac{513}{3}$
$\Rightarrow λ = - 171$

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

171

3511​

−171

513