Question

Let ${{a}_{1}},{{a}_{2}},.......$ and ${{b}_{1}},{{b}_{2}},........$ be the arithmetic progressions such that ${{a}_{1}}=25$, ${{b}_{1}}=75$ and ${{a}_{100}}+{{b}_{100}}=100$. The sum of the first hundred terms of the progression $\left( {{a}_{1}}+{{b}_{1}} \right),\left( {{a}_{2}}+{{b}_{2}} \right),..........$ is
(A) 0
(B) 100
(C) 10,000
(D) 5,05,000

Answer 1

Hint: We start solving this question by first adding the two progressions ${{a}_{1}},{{a}_{2}},.......$ and ${{b}_{1}},{{b}_{2}},........$, and then showing that the progression $\left( {{a}_{1}}+{{b}_{1}} \right),\left( {{a}_{2}}+{{b}_{2}} \right),..........$ is an arithmetic progression. Then we consider the formula for sum of n terms of A.P $\dfrac{n}{2}\left( 2{{a}_{1}}+\left( n-1 \right)d \right)$ and the formula for the ${{n}^{th}}$ term of A.P ${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$ and transform the formula for the sum of n terms into the terms of first and last term. Then we substitute the given values of ${{a}_{1}}$, ${{b}_{1}}$ and ${{a}_{100}}+{{b}_{100}}$ in the formula for n=100 and solve it to find the required value.

Complete step by step answer:
We are given that ${{a}_{1}},{{a}_{2}},.......$ and ${{b}_{1}},{{b}_{2}},........$ are two arithmetic progressions.
Let us assume that the common difference of the arithmetic progression ${{a}_{1}},{{a}_{2}},.......$ is ${{d}_{1}}$.
Then, we have
${{a}_{n+1}}-{{a}_{n}}={{d}_{1}}$
Let us assume that the common difference of the arithmetic progression ${{b}_{1}},{{b}_{2}},........$ is ${{d}_{2}}$.
Then, we have
${{b}_{n+1}}-{{b}_{n}}={{d}_{2}}$
Now let us consider the progression $\left( {{a}_{1}}+{{b}_{1}} \right),\left( {{a}_{2}}+{{b}_{2}} \right),..........$.
Let us consider the difference between two immediate terms in the above progression.
$\begin{align}
  & \Rightarrow \left( {{a}_{n+1}}+{{b}_{n+1}} \right)-\left( {{a}_{n}}+{{b}_{n}} \right) \\
 & \Rightarrow \left( {{a}_{n+1}}-{{a}_{n}} \right)+\left( {{b}_{n+1}}-{{b}_{n}} \right)={{d}_{1}}+{{d}_{2}} \\
\end{align}$
So, as the difference between the consecutive terms is constant, we can say that the progression $\left( {{a}_{1}}+{{b}_{1}} \right),\left( {{a}_{2}}+{{b}_{2}} \right),..........$ is an Arithmetic Progression.
Now we need to find the sum of first 100 terms of the progression $\left( {{a}_{1}}+{{b}_{1}} \right),\left( {{a}_{2}}+{{b}_{2}} \right),..........$.
Now let us consider the formula for sum of first n terms of an Arithmetic Progression ${{a}_{1}},{{a}_{2}},.......,{{a}_{n}}$ with common difference d.
$\dfrac{n}{2}\left( 2{{a}_{1}}+\left( n-1 \right)d \right)$
Now, let us also consider the formula for the ${{n}^{th}}$ term of the above AP.
${{a}_{n}}={{a}_{1}}+\left( n-1 \right)d$
Using this we can transform the formula for the sum of n terms of A.P as,
$\begin{align}
  & \Rightarrow \dfrac{n}{2}\left( 2{{a}_{1}}+{{a}_{n}}-{{a}_{1}} \right) \\
 & \Rightarrow \dfrac{n}{2}\left( {{a}_{1}}+{{a}_{n}} \right) \\
\end{align}$
So, using this formula for the sum of n terms of A.P, ${{S}_{n}}=\dfrac{n}{2}\left( {{a}_{1}}+{{a}_{n}} \right)$, we can find the sum of 100 terms of the progression $\left( {{a}_{1}}+{{b}_{1}} \right),\left( {{a}_{2}}+{{b}_{2}} \right),..........$ as,
$\Rightarrow {{S}_{100}}=\dfrac{n}{2}\left( \left( {{a}_{1}}+{{b}_{1}} \right)+\left( {{a}_{100}}+{{b}_{100}} \right) \right)$
As we are given that ${{a}_{1}}=25$, ${{b}_{1}}=75$ and ${{a}_{100}}+{{b}_{100}}=100$, let us substitute them in the above equation.
$\begin{align}
  & \Rightarrow {{S}_{100}}=\dfrac{100}{2}\left( \left( 25+75 \right)+\left( 100 \right) \right) \\
 & \Rightarrow {{S}_{100}}=50\left( 100+100 \right) \\
 & \Rightarrow {{S}_{100}}=50\left( 200 \right) \\
 & \Rightarrow {{S}_{100}}=10000 \\
\end{align}$

So, the correct answer is “Option C”.

Note: There is a possibility of one making a mistake while solving this problem by taking the formula for sum of n terms of A.P as, $\dfrac{n}{2}\left( {{a}_{1}}+\left( n-1 \right)d \right)$ which is equal to $\dfrac{n{{a}_{n}}}{2}$. But it is wrong. The actual formula for the sum of n terms of an A.P is $\dfrac{n}{2}\left( 2{{a}_{1}}+\left( n-1 \right)d \right)$.