If a1 + a2 + a3 + … + an = 3(2n+1 − 2), then a11 equals

Concept: If a sequence's partial sums are given by Sn = a1 + a2 + … + an, the individual term is recovered as an = Sn − Sn−1 (taking S0 = 0). If these differences follow a constant ratio, the sequence is a geometric progression (GP) whose nth term is a1·rn−1.

Application:

1. Write Sn = 3(2n+1 − 2), and note S0 = 3(21 − 2) = 0, so the formula for an = Sn − Sn−1 is valid from n = 1 onward.

2. Subtract: an = 3·2n+1 − 3·2n = 3·2n(2 − 1) = 3·2n.

3. Substitute n = 11: a11 = 3·211 = 3 × 2048 = 6144.

Cross-check: Compute the first three terms directly from Sn: a1 = S1 = 3(22 − 2) = 6; a2 = S2 − S1 = 3(23 − 2) − 6 = 12; a3 = S3 − S2 = 3(24 − 2) − 18 = 24. These form a GP with first term 6 and common ratio 2, so a11 = 6 × 210 = 6144 — matching the value from the general formula.