Question

Let $a_{1}, a_{2}, \ldots$ and $b_{1}, b_{2}, \ldots$ be arithmetic progression such that $a_{1}=25, b_{1}=75$ and $a_{100}+b_{100}=100,$ then the sum of first hundred term of the progression $a_{1}+b_{1}, a_{2}+b_{2}$, $\ldots$ is equal to

• A. 1000
• B. 100000
• C. 10000
• D. 24000

Answer 1

Answer: (C)

$\left(a_{100}+b_{100}\right)-\left(a_{1}+b_{1}\right)=99 d \Rightarrow d=0 $
$ \therefore $Sum of series$=(100 / 2)\left(a_{100}+b_{100}+a_{1}+b_{1}\right) $
$=50(100+100)=10000$