In an AP, it is given that, S₅ + S₇ =167 and S₁₀= 235, then find the AP, where S_n denotes the sum of its first n terms.

Let the first term be a.

Let Common difference be d.

Given: S₅ + S₇ =167

S₁₀= 235

Now, Sum of n terms of an arithmetic series is given by:

S_n = [2a + (n - 1)d]

So, consider

S₅ + S₇ =167

⇒ (5/2) [2a + (5 - 1)d] + (7/2) [2a + (7 - 1)d] = 167

⇒ (5/2) [2a + 4d] + (7/2) [2a + 6d] = 167

⇒ 5 × [a + 2d] + 7 × [a + 3d] = 167

⇒ 5a + 10d + 7a + 21d = 167

⇒ 12a + 31d = 167 …………….. (1)

Now, consider S₁₀= 235

⇒ (10/2) [2a + (10 - 1)d] = 235

⇒ 5 × [2a + 9d] = 235

⇒ 10a + 45d = 235

⇒ 2a + 9d = 47 ………. (2)

Subtracting equation (1) from 6 times of equation (2), we get,

⇒ 23d = 115

⇒ d = 5

So, from equation (2), we get,

a = 1/2 (47 - 9d)

⇒ a = 1/2 (47 - 45)

⇒ a = 1/2 (2)

⇒ a = 1

Therefore the AP is a, a + d, a + 2d, a + 3d,…

i.e. 1, 6, 11, 16, ….