Question

Find the number of terms of the. AP 54, 51, 48, ...... so that their sum is 513. Explain the double answer.

Answer 1

$a=54$

$d=51-54=-3$

$S_n=513$

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$\Rightarrow 513=\frac{n}{2}\left[2\right(54)+(n-1\left)\right(-3\left)\right]$

$\Rightarrow 1026=n[108-3n+3]$

$\Rightarrow 1026=n[111-3n]$

$\Rightarrow 3n^2-111n+1026=0$

$\Rightarrow n^2-37n+342=0$

$\Rightarrow n^2-18n-19n+342=0$

$\Rightarrow (n-18)(n-19)=0$

$\Rightarrow n=18,19$

$T_{18}=a+17d=54+17(-3)=3$

$T_{19}=a+18d=54+18(-3)=0$

Since the ${19}^{th}$ term is $0$, therefore no change will be seen in the sum of $18$ and $19$ terms. That is the reason why we are getting $2$ answers.