Question

Given is the image of a calendar. The sum of the numbers enclosed in the square is

A. 200
B. 170
C. 153
D. 270

Answer 1

Hint: As the numbers are in AP, having a common difference of 1, we will us the formula, ${{S}_{n}}=\dfrac{n}{2}\left\{ 2a+\left( n-1 \right)d \right\}$, where n is the number of terms in the AP, a is the first term of the AP and d is the common difference.

Complete step-by-step answer:
It is given in the question that we have to find the sum of the numbers enclosed in the square in the image of a calendar.

We can see that here; the numbers are in AP having a common difference of 1. We can also see that are 4 different Aps in the enclosed square. So, we will find the AP of each of them and find their sum to get the final answer.
So, the first AP is from 1 to 5. So, the sum of this AP will be given by, ${{S}_{n}}=\dfrac{n}{2}\left\{ 2a+\left( n-1 \right)d \right\}$, where n is the number of terms in the AP, a is the first term of the AP and d is the common difference. So, in this AP, we have n = 5, a = 1, d = 1. So, substituting these, we get the sum of the AP as,
$\begin{align}
  & {{S}_{n}}=\dfrac{5}{2}\left\{ 2\left( 1 \right)+\left( 5-1 \right)1 \right\} \\
 & \Rightarrow {{S}_{n}}=\dfrac{5}{2}\left\{ 2+4 \right\} \\
 & \Rightarrow {{S}_{n}}=\dfrac{5}{2}\times 6 \\
 & \Rightarrow {{S}_{n}}=5\times 3 \\
 & \Rightarrow {{S}_{n}}=15 \\
\end{align}$
Now, if we take the second AP, then it is from 8 to 12. So, we have the values of n = 5, a = 8 and d = 1. So, the sum of this AP will be,
$\begin{align}
  & {{S}_{n}}=\dfrac{5}{2}\left\{ 2\left( 8 \right)+\left( 5-1 \right)1 \right\} \\
 & \Rightarrow {{S}_{n}}=\dfrac{5}{2}\left\{ 16+4 \right\} \\
 & \Rightarrow {{S}_{n}}=\dfrac{5}{2}\times 20 \\
 & \Rightarrow {{S}_{n}}=5\times 10 \\
 & \Rightarrow {{S}_{n}}=50 \\
\end{align}$
So, for the third AP, we have the numbers from 15 to 19 and the values for n = 5, a = 15 and d = 1. So, we get the sum of the AP as,
$\begin{align}
  & {{S}_{n}}=\dfrac{5}{2}\left\{ 2\left( 15 \right)+\left( 5-1 \right)1 \right\} \\
 & \Rightarrow {{S}_{n}}=\dfrac{5}{2}\left\{ 30+4 \right\} \\
 & \Rightarrow {{S}_{n}}=\dfrac{5}{2}\times 34 \\
 & \Rightarrow {{S}_{n}}=5\times 17 \\
 & \Rightarrow {{S}_{n}}=85 \\
\end{align}$
And we have the last AP as from 22 to 26. SO, here n = 5, a = 22 and d = 1. So, the sum of the AP is,
$\begin{align}
  & {{S}_{n}}=\dfrac{5}{2}\left\{ 2\left( 22 \right)+\left( 5-1 \right)1 \right\} \\
 & \Rightarrow {{S}_{n}}=\dfrac{5}{2}\left\{ 44+4 \right\} \\
 & \Rightarrow {{S}_{n}}=\dfrac{5}{2}\times 48 \\
 & \Rightarrow {{S}_{n}}=5\times 24 \\
 & \Rightarrow {{S}_{n}}=120 \\
\end{align}$
Now, we will take the sum of all the obtained Aps to get our answer. So, we have the sum as,
$15+50+85+120=270$
Thus, the sum of the numbers under the enclosed square in the image will be 270.
Therefore, the correct option is option A.

Note: Many of the students try to solve this question by adding each of the numbers one by one to find the sum. But because of the complex addition of many numbers, they make calculation mistakes which would lead them to a wrong answer. So, it is always advisable to use the formula to find the sum without any errors.