Let a_{1}, a_{2}, a_{3}, ldots be a G.P. such | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $a_{1} + a_{2} = 4$ and $a_{3} + a_{4} = 16$.Since it is a $G$.$P$.,$a_{2} = a_{1}r$ and $a_{3} = a_{1}r^{2}$,$a_{4} = a_{1}r^{3}$.$a_{1}(1

Vedclass · Accepted Answer

$-171$

Vedclass · Answer

(A) Given $a_{1} + a_{2} = 4$ and $a_{3} + a_{4} = 16$.Since it is a $G$.$P$.,$a_{2} = a_{1}r$ and $a_{3} = a_{1}r^{2}$,$a_{4} = a_{1}r^{3}$.$a_{1}(1 + r) = 4$  --- $(1)$$a_{1}r^{2}(1 + r) = 16$ --- $(2)$Dividing $(2)$ by $(1)$,we get $r^{2} = 4$,so $r = 2$ or $r = -2$.If $r = 2$,$a_{1}(1 + 2) = 4 \Rightarrow a_{1} = 4/3 > 0$,which contradicts $a_{1} Thus,$r = -2$. Substituting in $(1)$,$a_{1}(1 - 2) = 4$ $\Rightarrow -a_{1} = 4$ $\Rightarrow a_{1} = -4$.The sum of the first $9$ terms is $S_{9} = a_{1} \frac{r^{9} - 1}{r - 1} = (-4) \frac{(-2)^{9} - 1}{-2 - 1} = (-4) \frac{-512 - 1}{-3} = (-4) \frac{-513}{-3} = (-4) 	imes 171 = -684$.Given $S_{9} = 4 \lambda$,we have $4 \lambda = -684 \Rightarrow \lambda = -171$.

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:

Similar Questions

The interior angles of an $n$-sided convex polygon are in $G.P.$ The smallest angle is $1^\circ$ and the common ratio is $2$. Then the number of possible values of $n$ is:

The three sides of a right-angled triangle are in $GP$ (geometric progression). If the two acute angles are $\alpha$ and $\beta$, then $\tan \alpha$ and $\tan \beta$ are

If in a geometric progression $\{a_n\}$,$a_1 = 3$,$a_n = 96$ and $S_n = 189$,then the value of $n$ is

If three geometric means are inserted between $2$ and $32$,then the third geometric mean will be

If the sum of $n$ terms of a $G.P.$ is $255$,the $n^{th}$ term is $128$,and the common ratio is $2$,then the first term will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module