The sum of how many terms of the A.P. 6, 12, 18, 24, ... is 1800?

The correct answer is Option (3) → 24

The terms 6, 12, 18, 24, … form an A.P. with first term \(a=6\) and common difference \(d=6\). The sum of the first \(n\) terms is \(S_n=\frac{n}{2}\left[2a+(n-1)d\right]\), which is set equal to 1800.

• Sum of the first \(n\) terms: \(\displaystyle S_n=\frac{n}{2}\left[12+6(n-1)\right]=\frac{n}{2}(6n+6)=3n(n+1)\)

• This sum must be 1800: \(3n(n+1)=1800\)

• So the product of \(n\) and the next number is \(n(n+1)=600\)

• \(600=24\times25\), two consecutive numbers, so \(n=24\)