Question

The sum of the first 16 terms of an AP whose first term and third term are 5 and 15 respectively is ?​

Answer 1

Answer:

a+a+2d=5+15

2(a+d)=20

a+d=10

5+d=10

d=5

and a=5.

s=(n/2)[2a+(n-1)d]

sum of first 16 term

s= 8[10 +75]

s=8*85=680

Answer 2

Answer:

1st method

$1st \: method \: \\ 1st \: term = 5 \\ 3rd \: term = 15 \\ then \: d = 5 \\ \\ 16term \: = a + 15d \\ = 5 + 15 \times 5 = 80 \\ sum = n \times \frac{a + l}{2} \\ 16 \times \frac{5 + 80}{2} \\ 16 + \frac{85}{2} \\ 8 \times 85 \\ = 680$

2nd method

$sum = no.of \: terms \: \times average \: of \: thar \: ap \\ sum \: = 16 \times \frac{5 + 80}{2} \\ 16 \times \frac{85}{2} \\ 8 \times 85 \\ = 680$

Step-by-step explanation:

Hope it helps you army :-)