The sum of the first `7` terms of an `AP` is `63` and the sum of its next `7 ` terms is `161.` Find the `28th` term of this `AP.`

The sum of the first `7` terms of an `AP` is `63` and the sum of its n

1.	The sum of the first `7` terms of an `AP` is `63` and the sum of its next `7 ` terms is `161.` Find the `28th` term of this `AP.`
Answer» Let a be the first term and d be the common difference of the given AP. Then, using `S_(n) = (n)/(2) * [2a + (n-1)d]`, we get `S_(7) = (7)/(2) (2a+6d) rArr 7(a+ 3d) = 63 " " [because S_(7) = 63]` `rArr a + 3d = 9. " "...(i)` Clearly, the sum of first 14 terms = 63 + 161 = 224. `therefore S_(14) = 224 rArr (14)/(2) (2a +13d) = 224` `rArr 7(2a +13d) = 224` `rArr 2a + 13d = 32. " "...(ii)` Multiplying (i) by 2 and subtracting the result from (ii), we get `7d = 14 rArr d = 2.` Putting d = 2 in (i), we get a = 9 -6=3. Thus, a = 3 and d =2. `therefore` 28th term of this AP is given by `T_(28) = (a+27d) = (3 + 27 xx 2) = 57.` Hence, the 28 th term of the given AP is 57.