Question

Given that $a_{4}+a_{8}+a_{12}+a_{16}=224,$ the sum of the first nineteen terms of the arithmetic progression $a_{1}, \, a_{2}, \, a_{3}, \, . \, \, . \, \, . \, \, . \, \, .$ is equal to

• A. $1540$
• B. $1064$
• C. $3125$
• D. $1980$

Answer 1

Answer: (B)

$a_{1},a_{2},a_{3},......,a_{19}$ are in $A.P.$
$\Rightarrow S=a_{1}+a_{2}+......+a_{18}+a_{19}=\frac{19}{2}\left(a_{1} + a_{19}\right)$
Given, $a_{4}+a_{8}+a_{12}+a_{16}=224$
$\Rightarrow a_{1}+3d+a_{1}+7d+a_{19}-7d+a_{19}-3d=224$
$\Rightarrow 2\left(a_{1} + a_{19}\right)=224$
$\Rightarrow a_{1}+a_{19}=112$
$\Rightarrow \frac{19}{2}\left(a_{1} + a_{19}\right)=\frac{19}{2}\times 112$
$\Rightarrow S=\frac{19}{2}\times 112=19\times 56=1064$