Question

Find the sum of first n term of A.P 6,2, -2....... .
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Answer 1

Answer:

4n-n^2

Step-by-step explanation:

Sn=n/2(2a+(n-1)d)

Sn=n/2(6*2+(n-1)-4)

   =n/2(12-4n+4)

   =n/2(16-4n)

   =n/2*2(8-2n)

   =n*2 (4-n)

   =2n*4 - 2n*n

   =8n -2n^2

   =4n -n^2

   HOPE IT HELPS U.....................

Answer 2

In the above A.P.

a=6

d= 2-6= -4

Now

$since \: sum \: of \: n \: terms \: = \frac{n(2a + (n - 1)d)}{2} \\ = \frac{n(2 \times 6 + (n - 1)( - 4))}{2} \\ = \frac{n(12 - 4n + 4)}{2} \\ = \frac{n(16 - 4n)}{2} \\ = \frac{4n(4 - n)}{2} \\ = 2n(4 - n)$

Hope it is helpful