Question

In an AP. It is given that `s_5 + s_7 = 167  and s_10 = 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 thee AP. Then,

`s_5 + s_7 = 167`

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

⇒ 5a + 10 d +7a + 21d = 167

⇒ 12a + 31d = 167                     ................... (1) 

Also, 

`s_10 = 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 =  5in (1), we get

12a+31 × 5 = 167

⇒ 12a + 155 = 167

⇒ 12a = 167 -155 = 12 

⇒ a= 1 

Hence, the AP is 1, 6, 11, 16,…….