Question

How many terms of the series 18 + 15 + 12 + ...... when added together will give 45 ?

Answer 1

Let no. of required term is n.
then...
Sn = $\frac{n}{2}\times [2a+(n-1\left)d\right]$
45 = $\frac{n}{2}\times [2\times 18+(n-1\left)\right(-3\left)\right]$
45 = $\frac{n}{2}\times [39-3n]$
90 = 39n - 3$n^2$
$3n^2-39n+90$=0
after solving...
n = 10 & 3