Question

the sum of first 16 term of the ap ; 10, 6,2, is​

Answer 1

Answer:

a = 10, d = 6 - 10 = -4

S16 = n/2 (2a + (n - 1)×d

= 16/2 (2×10 + 16-1 × -4)

= 8 (20 - 60)

= 8 × -40

= -320

Answer 2

Answer:

The formula for the some sum of first n terms of a AP is n/2(2a+(n-1)d

Here,

N= 16

A= 10

common differences (D)= 6-10=2-6= -4

So sum of first 16 Terms= 16/2(2-10+(16-1)-4)

= 8(20+15-4

= 8( 20-60)

= 8(-40)

= 320

Step-by-step explanation:

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