Question

Let $a_1,a_2,a_3\dots ,a_n$ be in A.P. and $a_3,a_5,a_8,b_1,b_2,b_3,\dots ,bn$ be in G.P. If $a_9=40,$ then

• A. $\sum _{i=1}^9{a}_i^2=6144$
• B. $\sum _{i=1}^{\infty }\frac{1}{b_i}=\frac{1}{18}$
• C. $\sum _{i=1}^9{a}_i^2=6278$
• D. $\sum _{i=1}^{\infty }\frac{1}{b_i}=\frac{1}{28}$

Answer 1

Answer: (A), (B)

$\sum _{i=1}^{\infty }\frac{1}{b_i}=\frac{1}{18}$

Given, $a_1,a_2,a_3,\dots ,a_n$ are in A.P.
Let $a_1=a$ and $a_2-a_1=d$

$a_3,a_5,a_8,b_1,b_2,b_3,\dots ,bn$ are in G.P.
$\Rightarrow {(a+4d)}^2=(a+2d)(a+7d)$
$\Rightarrow a=2d$
Given $a_9=40\Rightarrow a+8d=40$
$\Rightarrow a=8,d=4$
$\therefore a_1=8,a_2=12,a_3=16,\dots ,a_9=40$

Now, $\sum _{i=1}^9{a}_i^2=a_1^2+a_2^2+\cdots +a_9^2$
$=8^2+{12}^2+{16}^2+\cdots +{40}^2$
$=4^2(2^2+3^2+4^2+\cdots +{10}^2)$
$=16(1^2+2^2+3^2+\cdots +{10}^2-1^2)$
$=16(\frac{10\times 11\times 21}{6}-1)$
$=6144$

Let common ratio $=r$
$\Rightarrow r=\frac{3}{2}.$
Then, $b_1=54,b_2=81,\dots$
$\sum _{i=1}^{\infty }\frac{1}{b_i}=\frac{1}{b_1}+\frac{1}{b_2}+\cdots =\frac{1}{18}$