#### Question

Find the sum of the first 13 terms of the A.P: -6, 0, 6, 12,....

#### Solution

In the given problem, we need to find the sum of terms for different arithmetic progressions. So, here we use the following formula for the sum of *n* terms of an A.P.,

`S_n = n/2 [2a + (n - 1)d]`

Where; *a* = first term for the given A.P.

*d* = common difference of the given A.P.

*n *= number of terms

-6, 0, 6, 12,....To 13 terms

Common difference of the A.P. (d) = `a_2 - a_1`

= 0 - (-6)

= 6

Number of terms* *(*n*) = 13

First term for the given A.P. (*a*) = -6

So, using the formula we get,

`S_n = 13/2 [2(-6) + (13 - 1)(6)]`

`= (13/2)[-12 + (12)(6)]`

`= (13/2)[-12 + 72]`

`= (13/2)[60]`

= 390

Therefore, the sum of first 13 terms for the given A.P. is 390

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Solution Find the Sum of the First 13 Terms of the A.P : — 6, 0, 6, 12,.... Concept: Sum of First n Terms of an AP.