Question

plz help me to do this.Give answer fast before 4 :00​

Answer 1

Answer:

first AP:

6 , 10 , 14...........

a = 6

d = 10 - 6 = 4

n = 20

$S_{n}= \frac{n}{2}(2a + (n - 1)d)\\\\S_{20}= \frac{20}{2}(2*6 + (20 - 1)4)\\\\S_{20}= 10(2*6 + 19*4)\\\\S_{20}= 10(12 + 76)\\\\S_{20} = 10 * 88\\\\S_{20} = 880$

second AP:

15 , 19 , 23........

a = 15

d= 19 - 15 = 4

n = 20

$S_{n}= \frac{n}{2}(2a + (n - 1)d)\\\\S_{20}= \frac{20}{2}(2*15 + (20 - 1)4)\\\\S_{20}= 10(2*15 + 19*4)\\\\S_{20}= 10(30 + 76)\\\\S_{20} = 10 * 106\\\\S_{20} = 1060$

the difference between the sum of 20 terms of second AP & first AP

= 1060 - 880

= 180

I hope it helps......