Question

He sum of first n terms of three arithmetic progressions are s1, s2 and s3 respectively. the
first term of each a.p. is 1 and common differences are 1, 2 and 3 respectively. prove that s1 +
s3 = 2s2 solve the number value with out n.

Answer 1

Solution:-

For 1st A.P. : 1,2,3,....

Here, a = 1

d = 1

Sum to n terms, s1 = (n/2)[2+(n-1)]

or, s1 = n(n+1)/2

For 2nd A.P. : 1,3,5,...

Here, a = 1

d = 2

Sum to n terms, s2 = (n/2)[2+(n-1)×2]

or, s2 = n^2

For 3rd A.P. : 1,4,7,...

Here, a = 1

d = 3

Sum to n terms, s3 = (n/2)[2+(n-1)×3]

or, s3 = n(3n-1)/2

LHS = s1+s3 = (n^2+n+3n^2-n)/2

= 2n^2 = 2×s2 = RHS