The sum of first 15 terms of an ap is 112 and the sum of its next four terms is 518 find the ap
Answers
Correct Question
The sum of first 16 terms of an AP is 112 and the sum of its next fourteen terms is 518. Find the AP.
Solution
→ Sn = n/2 × [2a + (n - 1)d]
→ S16 = 16/2 × [2a + (16 - 1)d]
→ 112 = 8 × (2a + 15d)
→ 14 = 2a + 15d __(1)
Now S16 + S14 = S30
→ 112 + 518 = 630
→ S30 = 30/2 × [2a + (30 - 1)d]
→ 630 = 15 × (2a + 29d)
→ 42 = 2a + 29d __(2)
On doing eq(2) - eq(1)
→ 28 = 14d
→ d = 2
Hence 14 = 2a + 30
→ a = - 8
Hence first term is - 8 and common difference is 2.
Correct Question:
The sum of first 16 terms of an A.P is 112 and sum of next 14 terms is 518. Find The A.P.
Solution:
Given:
=> S16 = 112
=> Sum of next 14 term = 518.
According to question,
=> Sn = n/2 × [2a + (n - 1)d]
=> S16 = 16/2 × [2a + (16 - 1)d]
=> 112 = 8(2a + 15d)
=> 112/8 = 2a + 15d
=> 14 = 2a + 15d ..........[1]
sum of 30 terms = sum of 1st 16 term + sum of next 14 terms
=> S30 = 112 + 518
=> 30/2 [2a + (30 - 1)d] = 630
=> 2a + 29d = 42 .............[2]
Subtracting equation [1] from equation [2], we get
=> 14d = 28
=> d = 2
substitute d = 2 in equation [1], we get
=> 2a + 15 × 2 = 14
=> a = -8
Therefore AP = -8, -6, -4, -2,.......