Question

If the sum of first 4terms of an ap is 40 and that of first 14 terms is 280 . Find the sum of its first n terms

Answer 1

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)

pls mark brainliest

Answer 2

S4 = 4/2[2a+3d]

40 = 2[2a+3d]

20 = 2a + 3d

2a + 3d = 20     -------1

S14 = 14/2[2a+13d]

280*2/14 = 2a + 13d

40 = 2a + 13d

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

Subtracting eq 1 from eq 2

2a + 13d = 40

2a +  3d = 20

-    -         -

-----------------------

    10d = 20

       d = 20/10

       d = 2

Substituting d in eq 1

2a + 3(2) = 20

2a + 6 = 20

2a = 20-6

2a = 14

a = 14/2

a = 7

Sn = n/2 [ 2a + (n-1)d]

    = n/2 [ 2(7) + (n-1)2]

    = n/2 * 2[ 7 + (n-1)]

    = n[ 7 + n - 1]

    = 7n + n^2 -n

Sn= n^2 + 6n

        or

Sn=  n(n+6)