If the sum of first four terms of an AP is 40 and that of first 14 . . .

Question : If the sum of first four terms of an AP is 40 and that of first 14 terms is 280. Find the sum of the first n terms.

Doubt by Saumya

Solution :

S₄=40 (Given)

S₁₄=280 (Given)

Sn=?

We know,
S_n=n/2[2a+(n-1)d]

S₄=40
4/2[2a+(4-1)d]=40

2[2a+3d]=40

2a+3d=40/2

2a+3d=20 — (1)

S₁₄=280

14/2[2a+(14-1)d]=280

7[2a+13d]=280

2a+13d=280/7

2a+13d=40 — (2)

Solving equation (1) and (2)

2a+ 3d=20

2a+13d=40
- - -
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0 -10d=-20

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-10d=-20

d=-20/-10

d=2

Putting d=2 in equation (1)
2a+3(2)=20
2a+6=20

2a=20-6

2a=14

a=14/2

a=7

Now,
S_n=n/2[2a+(n-1)d]
S_n=n/2[2(7)+(n-1)2]
S_n=n/2[14+2n-2]
S_n=n/2[12+2n]
S_n=(n/2)×2[6+n]
S_n=n(n+6)

Class 10