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Sum of n terms of an A.P., S_n =

Where a=First term, d=Common difference, n=No. in terms

For n=4, S₄= (given)

⇒ 2a + 3d=20

⇒ 2a=20-3d ………….(1)

For n=14,S₁₄= (given)

⇒ 2a + 13d=40 ………..(2)

Putting the value of 2a in eq(2) from eq(1)

⇒ 20-3d + 13d=40

⇒ 20 + 10d=40

⇒ 10d=20

⇒ d=2

Putting value of d in eq(1), we get

⇒ 2a=20-3(2)=20-6=14

⇒ a=7

Sum of n terms, S_n =

S_n =

S_n =n[7 + n-1]

S_n =n(n + 6)=n² + 6n

Or

The first 30 positive integers divisible by 6 are 6,12,18,24…..

We see that these are in A.P.

With a=6 and d=12-6=6

Sum of n terms, S_n =

S₃₀=

S₃₀=15(186)=2790