In an A.P., $S_{5} + S_{7} = 167$ and $S_{10} = 235,$ then find the A.P., where $S_{n}$ denotes the sum of first n terms.

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Solution

$s_{10} = 235 s_{10} = \frac{10}{2} (2 a + (10 - 1) d] 235 = 5 (2 a + 9 d] 47 = 2 a + 9 d a = \frac{47 - 9 d}{2} s_{5} + s_{7} = 167 \frac{5}{2} [2 a + (5 - 1) d] + \frac{7}{2} [2 a + (7 - 1) d] = 167 \frac{5}{2} [2 a + 4 d] + \frac{7}{2} [2 a + 6 d] = 167 5 [a + 2 d] + 7 [a + 3 d] = 167 5 a + 10 d + 7 a + 21 d = 167 12 a + 31 d = 167 12 (\frac{47 - 9 d}{2}) + 31 d = 167 282 - 54 d + 31 d = 167 23 d = 115 d = 5 a = \frac{47 - 9 (5)}{2} = 1$
AP is $1, 6, 11, 16, . . . . .$

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