Question

Find the sum of 16 term of an AP :10,6,2

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Answer 1

Answer:

sum=n2(a+l)=162(10-50)=8×(-40)=-320.

Answer 2

Answer:

The answer is -320

Explanation:

sum of the first n terms of an AP = n/2[2a + (n-1)d]

Here, a=10, d=6-10=-4, n=16

so, putting values,

sum of 16 terms=

$16 \div 2(2 \times 10 + (16 - 1) \times ( - 4)) \\ = 8 \times (20 - 4 \times 15) \\ = 8 \times(20 - 60) \\ = 8 \times ( - 40) \\ = - 320$

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