Let a1, a2, a3,………. 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 $\sum_{i=1}^{9} a_{i}=4 \lambda$, then $\lambda$ is equal to:

  1. (1) $-513$

  2. (2) $-171$

  3. (3) 171

  4. (4) $\frac{511}{3}$


Correct Option: , 2

Solution:

Since, $a_{1}+a_{2}=4 \Rightarrow a_{1}+a_{1} r=4$ …(i)

$a_{3}+a_{4}=16 \Rightarrow a_{1} r^{2}+a_{1} r^{3}=16$…(ii)

From eqn. (i), $a_{1}=\frac{4}{1+r}$ and substituting the value of

$a_{1}$, in eqn (ii),

$\left(\frac{4}{1+r}\right)^{r^{2}}+\left(\frac{4}{1+r}\right)^{r^{3}}=16$

$\Rightarrow 4 r^{2}(1+r)=16(1+r)$

$\Rightarrow r=4 \quad \therefore \quad r=\pm 2$

$r=2, a_{1}(1+2)=4 \Rightarrow a_{1}=\frac{4}{3}$

$r=-2, a_{1}(1-2)=4 \Rightarrow a_{1}=-4$

$\sum_{i=1}^{a} a_{i}=\frac{a_{1}\left(r^{q}-1\right)}{r-1}=\frac{(-4)\left((-2)^{9}-1\right)}{-2-1}$

$=\frac{4}{3}(-513)=4 \lambda \quad \Rightarrow \quad \lambda=-171$

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