Question

Find the minimum number of terms of the AP 64, 60, 56,... so that their sum is 540.

Answer 1

We can see that

$d=60-64=-4a=64$

We know that ,

Sum of first terms,

$S_n=\frac{n}{2}(2a+(n-1\left)d\right)\Rightarrow 540=\frac{n}{2}(2\times 64+(n-1)\times (-4\left)\right)$

$\Rightarrow 540=64n-2n^2+2n$

$\Rightarrow n^2-33n+270=0$

$\Rightarrow n^2-15n-18n+270=0$

$\Rightarrow n(n-15)-18(n-15)=0$

$\Rightarrow (n-15)(n-18)=0$

$\Rightarrow n=15,18$

Since we need a minimum number of terms therefore $n=15$