The 4th term of an arithmetic series is 14 and the 6th term is 18. What is the sum of the first 8 terms.

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The nth term of an arithmetic series with first term a and common difference d is a + (n - 1)*d. The sum of the first n terms is (n/2)*(2a + (n -1)*d)

Here, the 4th term is 14 and the 6th term is 18.

a + 3d = 14 and a + 5d = 18

2d = 4

=> d = 2

a = 8

The sum of the first 8 terms is (8/2)*(2*8 + 7*2) = 4*30 = 120

The required sum is 120

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