CAT 2022 Slot 2 · Quant · Question 19

The progression starts at 3 and rises by 4 each time.

The sum of the first n terms is {A}_{n}=\frac{n}{2}\left( 2\times 3+\left( n-1 \right)\times 4 \right).

Simplifying inside the bracket, 6+4n-4=4n+2.

So {A}_{n}=\frac{n\left( 4n+2 \right)}{2}=n\left( 2n+1 \right)=2{n}^{2}+n.

Now sum this from n=1 to 25: ∑{A}_{n}=2∑{n}^{2}+∑n.

The sum of the first 25 squares is \frac{25\times 26\times 51}{6}=5525.

The sum of the first 25 integers is \frac{25\times 26}{2}=325.

So ∑{A}_{n}=2\times 5525+325=11050+325=11375.

Dividing by 25, \frac{11375}{25}=455 — the third option.

Shortcut. Dividing by 25 at the end means the answer is just the average of {A}_{1} through {A}_{25}. Since {A}_{n}=2{n}^{2}+n, that average is 2\times \overline{{n}^{2}}+\overline{n}=2\times 221+13=455, using \overline{{n}^{2}}=\frac{\left( n+1 \right)\left( 2n+1 \right)}{6}=221.