CAT 2022 Slot 1 · Quant · Question 18

The nth term of any sequence is the sum of the first n terms minus the sum of the first n-1 terms.

So {a}_{n}=\left( n+2{n}^{2}\right)-\left( \left( n-1 \right)+2{\left( n-1 \right)}^{2}\right).

Expanding the second bracket, \left( n-1 \right)+2\left({n}^{2}-2n+1 \right)=n-1+2{n}^{2}-4n+2.

That simplifies to 2{n}^{2}-3n+1.

Subtracting, {a}_{n}=\left( 2{n}^{2}+n \right)-\left( 2{n}^{2}-3n+1 \right)=4n-1.

As a check, the terms run 3,7,11,15,\ldots — an arithmetic progression with common difference 4, as expected.

For {a}_{n} to be divisible by 9, 4n-1 must be a multiple of 9.

Testing the options in order: n=4 gives 15, n=7 gives 27, n=8 gives 31, n=9 gives 35.

Only 27 is divisible by 9, so the smallest such n among the options is 7 — the third option.

Shortcut. A sum of the form A{n}^{2}+Bn always yields the term 2An+\left( B-A \right), so here {a}_{n}=4n-1 can be written down directly. Then 4n\equiv 1 modulo 9 gives n\equiv 7, since 4\times 7=28.