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, $a_4+a_8+a_{12}+a_{16}=224$
$\Rightarrow a_1+3d+a_1+7d+a_{19}-7d+a_{19}-3d=224$
$\Rightarrow 2\left(a_1+a_{19}\right)=224$
$\Rightarrow a_1+a_{19}=112$
$\Rightarrow \frac{19}{2}\left(a_1+a_{19}\right)=\frac{19}{2}\times 112$
$\Rightarrow S=\frac{19}{2}\times 112=19\times 56=1064$