Question

Find the number of terms of the AP $17,15,13,..........$ so that their sum is $72$. Explain the double answer.

Answer 1

Here $a=17$ & $d=-2$

$S_n=72$

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

$72=\frac{n}{2}\left[2\right(17)+(n-1\left)\right(-2\left)\right]$

$72=n[17+(n-1\left)\right(-1\left)\right]$

$72=n[17-n+1]$

$72=n(18-n)$

$72=18n-n^2$

$n^2-18n+72=0$

$n^2-12n-6n+72=0$

$n(n-12)-6(n-12)=0$

$(n-12)(n-6)=0$

$n=12$ or $n=6$

so the series will be

$17,15,13,11,9,7,5,3,1,-1,-3,-5$ or

$17,15,13,11,9,7$

here in the 12 terms series last 6 numbers will be zero. so the number of term will be double.