The 49th term of an arithmetic progression is 23 and 62nd term is 37. What is…

For an arithmetic progression (AP) of n terms, the sum is Sn = (n/2)(a1 + an). A useful property follows from this: if two term-positions p and q satisfy p + q = n + 1, then ap + aq is always equal to a1 + an — the two terms are symmetric about the middle of the sequence.

1. Here the series has n = 110 terms, and the given terms are the 49th and 62nd. Check p + q = 49 + 62 = 111, which equals n + 1 = 111 — so these two terms are exactly the symmetric pair.

2. By the property above, a49 + a62 = a1 + a110 = 23 + 37 = 60.

3. Apply the sum formula: S110 = (110/2)(a1 + a110) = 55 × 60 = 3300.

Cross-check with the direct two-equation method: writing a + 48d = 23 and a + 61d = 37 and adding them gives 2a + 109d = 60. Since S110 = (110/2)(2a + 109d), this again gives 55 × 60 = 3300 — confirming the result.

So the sum of the first 110 terms of the series is 3300.