Question

Write the formula for the sum of terms in AP.

Answer 1

Hint: In this question, we need to write the formula for the sum of terms in an arithmetic progression. For this, we will first suppose a general arithmetic progression having 'a' as the first term and 'd' as a common difference. Then we will try to find the sum of n terms using the assumed arithmetic progression.

Complete step by step answer:
Here we need to write the formula for the sum of terms in an arithmetic progression. Let us suppose that the first term of an arithmetic progression is 'a' and the common difference is 'd'. Then our arithmetic progression will look like this,
 $ a,a+d,a+2d,\ldots \ldots $ .
If we suppose that, this AP has n terms, then we know that, $ {{n}^{th}} $ term of an AP is given by, $ {{a}_{n}}=a+\left( n-1 \right)d $ so our AP becomes \[a,a+d,a+2d,\ldots \ldots ,a+\left( n-2 \right)d,a+\left( n-1 \right)d\].
Let us suppose that the last term of an AP is denoted by l i.e. \[a+\left( n-1 \right)d=l\] so the second last term becomes l-d and so on. Our series looks like this,
\[a,a+d,a+2d,\ldots \ldots ,\left( l-d \right),l\].
Let us denote its sum by $ {{S}_{n}} $ so,
\[{{S}_{n}}=a+a+d+a+2d+\ldots \ldots +l-d+l\cdots \cdots \cdots \left( 1 \right)\].
Similarly, if we reverse the series, the sum will remain the same, so,
\[{{S}_{n}}=l+l-d+l-2d+\ldots \ldots +a+d+a\cdots \cdots \cdots \left( 2 \right)\].
Let us add equation (1) and (2) we get,
\[{{S}_{n}}+{{S}_{n}}=\left( a+l \right)+\left( a+d+l-d \right)+\left( a+2d+l-2d \right)+\ldots \ldots +\left( l+a \right)\].
Simplifying we get,
 $ 2{{S}_{n}}=\underset{\text{n times}}{\mathop{\left( a+l \right)+\left( a+l \right)+\left( a+l \right)+\ldots \ldots +\left( a+l \right)}}\, $ .
So adding same terms n times we get,
 $ 2{{S}_{n}}=n\left( a+l \right) $ .
Dividing both sides by 2, we get,
 $ {{S}_{n}}=\dfrac{n}{2}\left( a+l \right) $ .
Putting the value of l as $ a+\left( n-1 \right)d $ in the above formula we get,
 $ {{S}_{n}}=\dfrac{n}{2}\left( a+a+\left( n-1 \right)d \right) $ .
Simplifying we get,
 $ {{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right) $ .
This is the required formula for the sum of n terms in an arithmetic progression.

Note:
 Students should note that series either taken from the last term to the first term or from the first term to the last term, the sum remains the same. In case, we are just given the first term and the last term of an AP, we can directly use the formula $ {{S}_{n}}=\dfrac{n}{2}\left( a+l \right) $ to find the sum of n terms where l is last term ( $ {{n}^{th}} $ term).