Question

How many terms of the sequence 18,16,14 should be taken so that their sum is zero.
[a] 19
[b] 17
[c] 18
[d] 16

Answer 1

Hint: Assume that the number of terms is n. Use the formula ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$, where a is the first term of the A.P and d is the common difference of the A.P, to find the sum of the first n terms of the A.P.
Equate this to 0 and hence form a quadratic in n. Remove the extraneous roots(if any) and hence find the value of n. Verify your answer.

Complete step-by-step answer:
Let the sum of first n terms be n.
We know that the sum of first n terms of an A.P is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$, where a and d have their usual meanings.
Here a = 18, d = 16-18 = -2.
Hence we have
${{S}_{n}}=\dfrac{n}{2}\left( 2\times 18+\left( n-1 \right)\left( -2 \right) \right)=\dfrac{n}{2}\times 2\left( 18-n+1 \right)=n\left( 19-n \right)=19n-{{n}^{2}}$
But since the sum of first n terms is 0, we have
$19n-{{n}^{2}}=0$
Hence we have
$\begin{aligned} & {{n}^{2}}-19n=0 \\\ & \Rightarrow n\left( n-19 \right)=0 \\\ \end{aligned}$
Using zero product property, we have
$n=0$ or $n=19$
Since n is a natural number, we have $n>0$
Hence n = 0 is rejected.
Hence we have
n=19.
Hence we must take the first 19 terms of the A.P so that their sum is 0.
Hence option [a] is correct.

Note: [1] Verification:
The sum of the first 19 terms is given by ${{S}_{19}}=\dfrac{19}{2}\left( 2\times 18+18\left( -2 \right) \right)=\dfrac{19}{2}\left( 36-36 \right)=0$
Hence the sum of the first 19 terms is 0.
Hence the answer is verified.
[2] Alternatively, we have
If the sum of first n terms is 0, then $n=\dfrac{-2a}{d}+1$
Here a = 18 and d = -2
Hence we have
$n=\dfrac{-2\times 18}{-2}+1=18+1=19$, which is the same as obtained above.