Question

How many terms of the A.P,$3,5,7,9,-----$ must be added to get the sum of $120?$

Answer 1

$120=\frac{n}{2}[a+a+(n-1\left)d\right]$

where $n=$ no of terms

$a=$first term

$d=$common difference

$120=\frac{n}{2}[3+3+(n-1\left)2\right]240=n[3+3+(n-1\left)2\right]240=n[6+2n-2]240=n[4+2n]n^2+2n-120=0n^2+12n-10n-120=0n(n+12)-10(n+12)=0(n-10)(n+12)=0n=10$

no of terms must be added $10-4=6$ terms