In an AP of 50 terms, the sum of first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the AP.

Given that, a number of terms in AP = n = 50.

We know that sum of first n terms of an AP,

Where a = first term, d = common difference.

Now,

S₁₀ = 10[2a + (10 – 1) d]/2

But given, S₁₀ = 210

So,

⇒ 10[2a + (10 – 1) d]/2 = 210

⇒ 2a + 9d = 210/5

⇒ 2a + 9d = 42 … (1)

Given, sum of last 15 terms = 2565

Sum of first 50 terms – Sum of first 35 terms = 2565

S₅₀ – S₃₅ = 2565

⇒ 25[2a + 49d] – 35[2a + 34d]/2 = 2565

Dividing by 5 on both sides,

⇒ 5[2a + 49d] – 7[2a + 34d]/2 = 513

⇒ 10a + 245d – 7a – 119d = 513

⇒ 3a + 126d = 513

⇒ a + 42d = 171 … (2)

Multiplying (2) with 2, we get

⇒ 2a + 84d = 342 … (3)

Subtracting (1) from (3), we get

⇒ 75d = 300

⇒ d = 300/75

⇒ d = 4

Substituting d value in (1),

⇒ 2a + 9(4) = 42

⇒ 2a = 6

⇒ a = 3

We know that the general form of AP is a, a + d, a + 2d, a + 3d …

∴ AP = 3, 3 + 4, 3 + 2(4), 3 + 3(4) …

⇒ AP = 3, 7, 11, 15 …

Ans. The AP formed by the given conditions is 3, 7, 11, 15 …