Question

How many terms of the A.P. 18, 16, 14, …. be taken, so that their sum is zero.

Answer 1

Hint: We first try to find the general equations and series for an A.P. We assume the first term and the common difference. We find the equation of general term and sum of first n terms of the series. Then we put the values of 2n, 3n and 5n in place of the equations and find the relation between ${{S}_{2}},{{S}_{3}},{{S}_{5}}$.

Complete step by step answer:
We express the A.P. in its general form.
We express the terms as ${{t}_{n}}$, the ${{n}^{th}}$ term of the series.
The first term is 18. So, ${{t}_{1}}=18$. The common difference is d. So, $d=16-18=14-16=-2$.
We express general term ${{t}_{n}}$ as ${{t}_{n}}=18+\left( n-1 \right)\left( -2 \right)=18+2-2n=20-2n$.
The sum of n terms will be ${{S}_{n}}$ as ${{S}_{n}}=\dfrac{n}{2}\left[ 2\times 18+\left( n-1 \right)\left( -2 \right) \right]=\dfrac{n}{2}\left[ 36+2-2n \right]=n\left( 19-n \right)$.
We need to find the number of terms of the A.P. 18, 16, 14, …… for which ${{S}_{n}}$ becomes 0.
So, we put the equation ${{S}_{n}}=n\left( 19-n \right)=0$. Solving the equation, we get $n=0,19$.
Value of n can’t be 0. So, the value of n is 19.

We have to take 19 terms of the A.P. 18, 16, 14, …. To make the sum 0.

Note: The series is actually a descending series with difference -2. After a certain number of iterations, the series reaches 0. Then negative terms start to appear. The sum of these positive and negative terms becomes 0.