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=19a_i=4λ, then λ, is equal to.

Correct answer: (B) -171. a_1+a_2=4⇒a_1+a_1r=4…1 a_3+a_4=16⇒a_1r^2+a_1r^3=16…2 (1r^2=1)/4⇒r^2=4 ⇒r=-2 a_1<0 ∑i=1aa_i=(a_1r^9-1)/(r-1)=(-4-2^9-1)/(-2-1)=4/3-513=4λ λ=-171.

Q28 / 75 · JEE Main 2020 (07 Jan Shift 2) · Mathematics

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 $\sum_{i = 1}^{9} a_{i} = 4 λ$ , then $λ$ , is equal to.

A
$- 513$
Correct answer
B
$- 171$
C
$171$
D
$\frac{511}{3}$

Solution

$a_{1} + a_{2} = 4 \Rightarrow a_{1} + a_{1} r = 4 \dots (1)$

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

$\frac{1}{r^{2}} = \frac{1}{4} \Rightarrow r^{2} = 4 \Rightarrow r = - 2 (a_{1} < 0)$

$\sum_{i = 1}^{a} a_{i} = \frac{a_{1} (r^{9} - 1)}{r - 1} = \frac{(- 4) [{(- 2)}^{9} - 1]}{(- 2 - 1)} = \frac{4}{3} (- 513) = 4 λ$

$λ = - 171$ .

Hence, the correct answer is option: 'B'.

Let a1,a2,a3,…, be a G.P. such that a1<0,a1+a2=4 and a3+a4=16. If ∑i=19ai=4λ, then λ, is equal to.

Solution

Relevant Questions

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 $\sum_{i = 1}^{9} a_{i} = 4 λ$ , then $λ$ , is equal to.