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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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