Let a_1,a_2,a_3, be a G.P.

Question

Let $a_1,a_2,a_3,\ldots$ be a G.P. such that $a_1<0$, $a_1+a_2=4$, and $a_3+a_4=16$. If $\displaystyle\sum_{i=1}^{9}a_i = 4\lambda$, then $\lambda$ is equal to:

Accepted Answer

${a_1}$
 + ${a_2}$
 = 4

$$ \Rightarrow $$ ${a_1}$
 + ${a_1}$r
 = 4 ...(1)

${a_3}$
 + ${a_4}$
 = 16

$$ \Rightarrow $$ ${a_1}$r2
 + ${a_1}$r3
 = 16 ...(2)

Doing (1) $$ \div $$ (2), we get

r = $$ \pm $$ 2

If r = 2, then a1 = $${4 \over 3}$$

If r = -2, then a1 = -4

Given ${a_1}$
 < 0

$$ \therefore $$ a1 = -4

$$ \therefore $$  $$\sum\limits_{i = 1}^9 {{a_i}}$$ = $${{a\left( {{r^9} - 1} \right)} \over {r - 1}}$$ = 4$$\lambda $$

$$ \Rightarrow $$ $${{ - 4\left( {{{\left( { - 2} \right)}^9} - 1} \right)} \over { - 2 - 1}}$$ = 4$$\lambda $$

$$ \Rightarrow $$ $$\lambda $$ = -171

Solutions · 1