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Question
In an A.P., if S5 + S7 = 167 and S10=235, then find the A.P., where Sn denotes the sum of its first n terms.
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 ,..........
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