Question

If the $56^{th}$ term of an AP is $\dfrac{10}{37}$, find the sum of the first 111 terms.

Answer 1

Here, we have been given the $56^{th}$ term of an AP and we have to find the sum of the first 111 terms of the respective AP. For this, we will first write the given term, i.e. the $56^{th}$ term in the form of the first term ‘a’ and common difference ‘d’ by using the formula ${{a}_{n}}=a+\left( n-1 \right)d$. Then we will write the sum of the first 111 terms in terms of ‘a’ and ‘d’ too using the formula $Sum=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$. Then we will compare the 2 equations thus formed and hence we will get the value of the sum. Thus, we will get our required answer.

Complete step by step answer:
Here, we have been given the $56^{th}$ term of an AP to be $\dfrac{10}{37}$. Now, we know that the $n^{th}$ term of an AP with first term ‘a’ and common difference ‘d’ is given as:
${{a}_{n}}=a+\left( n-1 \right)d$
Here, we have:
$\begin{aligned} & n=56 \\\ & {{a}_{n}}=\dfrac{10}{37} \\\ \end{aligned}$
Thus, for this AP, we get:
$\dfrac{10}{37}=a+\left( 56-1 \right)d$
$\Rightarrow \dfrac{10}{37}=a+55d$ …..(i)
Now, we have to find the sum of the first 111 terms of this AP.
We know that the sum of ‘n’ terms of an AP is given as:
$Sum=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$
Here, we have:
$n=111$
Thus, the sum will be:
$\begin{aligned} & Sum=\dfrac{111}{2}\left( 2a+\left( 111-1 \right)d \right) \\\ & \Rightarrow Sum=\dfrac{111}{2}\left( 2a+110d \right) \\\ \end{aligned}$
Now, simplifying it we get:
$\begin{aligned} & Sum=\dfrac{111}{2}\left( 2a+110d \right) \\\ & \Rightarrow Sum=\dfrac{111}{2}\left( 2\left( a+55d \right) \right) \\\ \end{aligned}$
$\Rightarrow Sum=111\left( a+55d \right)$ …..(ii)
Now, putting the value of equation (i) in equation (ii), we get:
$\begin{aligned} & Sum=111\left( a+55d \right) \\\ & \Rightarrow Sum=111\left( \dfrac{10}{37} \right) \\\ \end{aligned}$
Solving it we get:
$\begin{aligned} & Sum=111\left( \dfrac{10}{37} \right) \\\ & \Rightarrow Sum=3\left( 10 \right) \\\ & \therefore Sum=30 \\\ \end{aligned}$

Thus, the required sum is 30.

Note: We could have also divided equations (i) and (ii) to obtain the answer. This is shown as follows:
$\begin{aligned} & \dfrac{equation\left( ii \right)}{equation\left( i \right)} \\\ & \Rightarrow \dfrac{Sum}{\dfrac{10}{37}}=\dfrac{111\left( a+55d \right)}{a+55d} \\\ & \Rightarrow \dfrac{Sum}{\dfrac{10}{37}}=111 \\\ & \Rightarrow Sum=111\left( \dfrac{10}{37} \right) \\\ & \therefore Sum=30 \\\ \end{aligned}$