Sum

In an A.P., if S5 + S7 = 167 and S_{10=}235, then find the A.P., where Sn denotes the sum of its first n terms.

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#### Solution

`"S"_5+ "S"_7= 167 and "S"_10=235`

Now `"S"_n=n/2[ 2a + (n-1) d ]`

`"S"_5 + "S"_7=167`

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

⇒ 12a + 31d = 167 .......(i)

also `"S"_10=235`

∴ `10/2 [ 2a + 9d ] = 235`

2a + 9d = 47 .........(ii)

Multiplying equation (2) by 6, we get

12a + 54d = 282 .....(3)

(-) 12a + 31d = 167

- - -

23 d = 115

`therefore d = 5`

Substituting value of d in (2), we have

2a + 9(5) = 47

2a + 45 = 47

2a = 2

a = 1

Thus, the given A.P. is 1, 6, 11, 16 ,..........

Concept: Sum of First n Terms of an AP

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