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Question:

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

Solution:

Let the first term be a and the common difference be d.
By using Sn = n/2[2a + (n-1)d], we have,
S5 = 5/2[2a + (5-1)d] = 5/2[2a + 4d]
S7 = 7/2[2a + (7-1)d] = 7/2[2a + 6d]
Given: S7 + S5 = 167
∴ 5/2[2a + 4d] + 7/2[2a + 6d] = 167
⇒ 10a + 20d + 14a + 42d = 334
⇒ 24a + 62d = 334.. (1)
S10 = 10/2[2a + (10-1)d] = 5(2a + 9d)
Given: S10 = 235
So 5(2a + 9d) = 235
⇒ 2a + 9d = 47.. (2)
Multiply equation (2) by 12, we get
24a + 108d = 564 (3)
Subtracting equation (3) from (1), we get
-46d = -230
∴ d = 5
Substituting the value of d = 5 in equation (1) we get
2a + 9(5) = 47
or 2a = 2
∴ a = 1
Then A.P is 1, 6, 11, 16, 21,…