Question

If ${{a}_{1}},{{a}_{2}},{{a}_{3}}........$ are in A.P. such that ${{a}_{1}}+{{a}_{7}}+{{a}_{16}}=40$ , then the sum of the first $15$ terms of this A.P. is:
A. 200
B. 280
C. 120
D. 150

Answer 1

Hint: We are given the arithmetic progression sequence, we will apply the conventional formula for ${{n}^{th}}$ term that is , ${{a}_{n}}=a+\left( n-1 \right)d$ and then we will find one equation in terms of a and d. Then we apply the formula for sum of n terms for an arithmetic progression which is ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$ and then we will put the value from the equation obtained earlier in terms of a and d.

Complete step by step answer:
Since it is given that the sequence ${{a}_{1}},{{a}_{2}},{{a}_{3}},.......,{{a}_{n}},....$ is in arithmetic progression, therefore it will be in the following form:
$a,a+d,a+2d,.......,a+\left( n-1 \right)d,.....$
Where, ${{a}_{n}}=a+\left( n-1 \right)d$ is the ${{n}^{th}}$ term and $a$ is the first term in the sequence and $d$ is the common difference between terms.
Let’s take the first equation given in the equation that is: ${{a}_{1}}+{{a}_{7}}+{{a}_{16}}=40\text{ }.........\text{Equation 1}\text{.}$ We will first find the values of these terms in terms of $a$ and $d$ :
We know that: ${{a}_{n}}=a+\left( n-1 \right)d$ .
Therefore,
$\begin{align}
  & {{a}_{1}}=a+\left( 1-1 \right)d\Rightarrow {{a}_{1}}=a \\
 & {{a}_{7}}=a+\left( 7-1 \right)d\Rightarrow {{a}_{7}}=a+6d \\
 & {{a}_{16}}=a+\left( 16-1 \right)d\Rightarrow {{a}_{16}}=a+15d \\
\end{align}$

Now we will put these values in equation 1:
 $\begin{align}
  & {{a}_{1}}+{{a}_{7}}+{{a}_{16}}=40 \\
 & \Rightarrow a+\left( a+6d \right)+\left( a+15d \right)=40 \\
 & \Rightarrow 3a+21d=40\Rightarrow 3\left( a+7d \right)\text{=40} \\
 & \Rightarrow a+7d=\dfrac{40}{3}\text{ }...........\text{ Equation 2}\text{.} \\
\end{align}$
 Now we know that the sum of an arithmetic progression is: ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$ .
Now we have to find out the sum of first 15 terms of the A.P. , therefore, we will take $n=15$
\[\begin{align}
  & \Rightarrow {{S}_{n}}=\dfrac{15}{2}\left( 2a+\left( 15-1 \right)d \right)=\dfrac{15}{2}\left( 2a+14d \right)=\dfrac{15}{2}\times 2\left( a+7d \right) \\
 & \Rightarrow {{S}_{n}}=15\left( a+7d \right) \\
\end{align}\]
Now we will be putting the value of $a+7d$ from equation 3, therefore: $\begin{align}
  & {{S}_{n}}=15\times \dfrac{40}{3}=5\times 40=200 \\
 & \Rightarrow {{S}_{n}}=200 \\
\end{align}$
Therefore, the sum of terms of the first 15 terms in the given arithmetic progression is $200$.

So, the correct answer is “Option A”.

Note: In questions like these, since only one condition is given you may not be able to find the values of $a$ and $d$ , as only one equation will be there. So when you find out the sum of the terms using the formula ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$ try and look out if you can fit the relation between $a$ and $d$ here, because in this way you will be able to get your answer.