Question

In an AP, it is given that $S_5+S_7=167and{S}_{10}=235,$ 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.

$S_5$ +$S_7$ = 167

⇒ $\frac{5}{2}$[2a + (5 - 1)d] + $\frac{7}{2}$[2a + (7 - 1)d] = 167

⇒ 12a + 31d =167 ------(i)

$S_{1}0$ = 235

⇒ $\frac{10}{2}$[2a + (10 - 1)d] = 235

⇒ 10a +45d =235

⇒ 2a + 9d =47 -------(ii)

Multiply (ii) by 6, we get;

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

Subtracting (i) from (iii),

we get 54d − 31d = 282−167

⇒23d = 115

⇒d = 5

Now, from (ii), we get 2a + 9 x 5 = 47

⇒2a = 2

⇒a = 1

Now, the AP is:

,a, a + d, a+2d, a+3d.. ........

1, 6, 11, 16 ......