Question

If the sum of first 4 terms of an AP is 40 and that of first 14 the is 280, find the sum of first n term

Answer 1

Sn = sum of n terms of an A.P. = n/2 [ 2a + (n-1)d]

A/Q   

 S₄ = 40 = 4/2 [2a + 3d]  =  2a + 3d = 20            ...........(1)

and


S₁₄ = 280 = 14/2 [2a + 13d]  = 2a + 13d = 40         .........(2)

solving (1) and (2)

gives  a = 7    and d= 2

so Sn = n/2[ 2(7) + (n-1)2]  = 7n + n² - n = n² + 6n

hope my solution is true.

Answer 2

S4 = 40  

2(2a + 3d) = 40  

2a + 3d = 20-------1

S14 = 280  

7(2a + 13d) =280  

2a + 13d = 40-----2

From 1 and 2 eq. we get

Solving WE to get d = 2

and a = 7

∴Sn=n2[14+(n−1)×2]

= n(n + 6) or (n2 + 6n)