Question

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.

Answer 1

Let a be the first term and d be the common difference of the AP. Then,

S₅ + S₇ = 167

⇒ `5/2 (2a + 4d) + 7/2 (2a + 6d) = 167`   ...`{S_n = n/2 [2a + (n - 1)d]}`

⇒ 5a + 10d + 7a + 21d = 167

⇒ 12a + 31d = 167   ...(1) 

Also, 

S₁₀ = 235

⇒ `10/2 (2a + 9d) = 235`

⇒ 5(2a + 9d) = 235

⇒ 2a + 9d = 47

Multiplying both sides by 6, we get

12a + 54d = 282   ...(2)

Subtracting (1) from (2), we get

12a + 54d – 12a – 31d = 282 – 167

⇒ 23d = 115

⇒ d = 5

Putting d = 5 in (1), we get

12a + 31 × 5 = 167

⇒ 12a + 155 = 167

⇒ 12a = 167 – 155

⇒ 12a = 12 

⇒ a = 1 

Hence, the AP is 1, 6, 11, 16,.......