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

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Solution

There are two formulae to find the sum of the terms of an $A P$ .
In this case we know the following:
$a = - 6, d = 6, n = 13$
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
$S_{13} = \frac{13}{2} [2 (- 6) + (12) 4]$
$S_{13} = 6.5 [36]$
$S_{13} = 234$ .

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

There are two formulae to find the sum of the terms of an AP.In this case we know the following:a=−6,d=6,n=13Sn=n2[2a+(n−1)d]S13=132[2(−6)+(12)4]S13=6.5[36]S13=234.

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

There are two formulae to find the sum of the terms of an $A P$ .
In this case we know the following:
$a = - 6, d = 6, n = 13$
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
$S_{13} = \frac{13}{2} [2 (- 6) + (12) 4]$
$S_{13} = 6.5 [36]$
$S_{13} = 234$ .