Question

If the first term of an A.P. is 61 and the second term is 57. Then find the value of n such the sum of first n terms is equal to n.

Answer 1

Hint: We already know the first term and can find out the common difference by subtracting the first term from the second. So, we just have to use the formula for the sum of first n terms of an A.P. and equate the sum with n and solve the equation to reach the answer.

Complete step-by-step answer:
Before starting with the solution, let us discuss what an A.P. is. A.P. stands for arithmetic progression and is defined as a sequence of numbers for which the difference of two consecutive terms is constant. The general term of an arithmetic progression is denoted by ${{T}_{n}}$ and sum till n terms is denoted by ${{S}_{n}}$ .
${{T}_{n}}={{a}_{1}}+\left( n-1 \right)d$
${{S}_{n}}=\dfrac{n}{2}\left( 2{{a}_{1}}+\left( n-1 \right)d \right)=\dfrac{n}{2}\left( {{a}_{1}}+{{a}_{n}} \right)$
Now moving to the solution of the question, it is given ${{a}_{1}}=61$ and we know $d={{a}_{2}}-{{a}_{1}}= 61 - 57 = -4$ . Therefore, the sum of first n terms of the given A.P. is:
${{S}_{n}}=\dfrac{n}{2}\left( 2{{a}_{1}}+\left( n-1 \right)d \right)=\dfrac{n}{2}\left( 2\times 61+\left( n-1 \right)\left( -4 \right) \right)$
Now it is given that the sum of the first n terms is equal to n. Therefore, we get
$n=\dfrac{n}{2}\left( 2\times 61-\left( n-1 \right)4 \right)$
We know that n is not zero, so cancelling n from both sides, we get
$1=\dfrac{1}{2}\left( 122-\left( n-1 \right)4 \right)$
$\Rightarrow 2=122-\left( n-1 \right)4$
$\Rightarrow 120=\left( n-1 \right)4$
$\Rightarrow 30=n-1$
$\Rightarrow n=31$
Therefore, the value of n is 31.

Note: Always, while dealing with an arithmetic progression, try to find the first term and the common difference of the sequence, as once you have these quantities, you can easily solve the questions related to the given arithmetic progressions.