Question

Hello there
Question 2
Pls ans

Answer 1

Sum of first 20 terms:

a = 6

d = 4

n = 20

$\sf{s_{n} = \frac{n}{2}(2a + (n-1)d}$

=> $\sf{s_{20} = \frac{20}{2}(2(6) + (20-1)4}$

=> $\sf{s_{20} = \frac{20}{2}(12 + 76}$

=> Sum of first 20 terms = 880

Sum of next 20 terms:

a = 21st term of the AP

a = a + (n - 1)d = 6 + 20(4)

a = 86

d = 4

n = 20

$\sf{s_{n} = \frac{n}{2}(2a + (n-1)d}$

=> $\sf{s_{20} = \frac{20}{2}(2(86) + (20-1)4}$

=> $\sf{s_{20} = \frac{20}{2}(172 + 76}$

=> Sum of next 20 terms = 2480

Difference = 2480 - 880

Difference = 1600