Question

In an AP of 50 terms, the sum of first 10 terms is 2 and the sum of last 15 terms is 2565. Find the AP.​

Answer 1

In an A.P. of 50 terms, we are given:

• Sum of 10 terms = 210
• Sum of 15 terms = 2565

→ Sₙ = n/2 × [2a + (n - 1)d]

→ S₁₀ = 10/2 × [2a + (10 - 1)d]

→ 210 = 5[2a + 9d]

→ 42 = 2a + 9d ____(1)

→ S₅₀ - S₃₅ = S₁₅

→ 50/2 [2a + 49d] - 35/2 [2a + 34d = 2565

→ 5 [2a + 49d] - 7/2 [2a + 34d] = 513

→ 10a + 245d - 7a - 139d = 513

→ 3a + 126d = 513

→ a + 42d = 171 ___(2)

By doing 2 eq(2) - eq(1),

2a + 84d = 342

- (2a + 9d = 42)

—————————

→ 75d = 300

—————————

→ d = 300/75

→ d = 4

→ a = 171 - 42d

→ a = 171 - 168

→ a = 3

So A.P. we attained is,

→ 3, 7, 11, 15, 19, 23, 27.......