Question

the sums of first n terms of 3ap's are s1,s2 and s3. the first term of each is 5 and there common differences are 2,4 and 6 respectively prove that s1+ s3=2s2.​

Answer 1

Answer:

s1+ s3=2s2.​

Step-by-step explanation:

the sums of first n terms of 3ap's are s1,s2 and s3

First Term of each AP = 5

common differences are 2,4 and 6

Sum of n terms of an AP = (n/2)(a + a + (n-1)d)

= (n/2)(2a + (n-1)d)

a = 5  d = 2 , 4 , 6

s1 = (n/2)(2*5 + (n-1)2)  = n(5 + n-1)  = n(n + 4)

s2 = (n/2)(2*5 + (n-1)4)  = n(5 + 2n-2)  = n(2n + 3)

s3 = (n/2)(2*5 + (n-1)6)  = n(5 + 3n-3)  = n(3n + 2)

s1 + s3 = n(n + 4) +  n(3n + 2)

= n(n + 4 + 3n + 2)

=n(4n + 6)

= 2n(2n + 3)

= 2s2

Hence Proved

s1+ s3=2s2.​

Answer 2

Answer:

272+332+53+3+3+345 =2873729

Hope helps mate