Question

Find the number of terms of the AP $64,60,56$,___ so that their sum is $544$. Explain the double answer.

Answer 1

Here $a=64$ , $d=-4$ & $S_n=544$

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

$544=\frac{n}{2}\left[2\right(64)+(n-1\left)\right(-4\left)\right]$

$544=n(64-(n-1\left)\right(-2\left)\right)$

$544=n(64-2n+2)$

$544=n(66-2n)$

$544=66n-2n^2$

$2n^2-66n+544=0$

$n^2-33n+272=0$

$n^2-17n-16n+272=0$

$n^2-17n-16n+272=0$

$n(n-17)-16(n-17)=0$

$(n-16)(n-17)=0$

here we get two answer because last terms is zero

as ${17}^{th}$ term $=a+16d$

$=64+16(-4)=0$

so the sum of first $16$ and $17$ number will be same