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 its firstn terms
36.​

Answer 1

Given

We have given S₄= 40 and S₁₄= 280

To find

We have to find the sum of its nth term

$\sf\huge\bold{\underline{\underline{{Solution}}}}$

We will apply sum formula

sₙ= n/2 [2a +(n-1)d]

⇢s₄= 4/2(2a+(4-1)d

⇢s₄= 2(2a+3d)

⇢40= 2(2a+3d)

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

Now,

⇢S₁₄= 14/2(2a+(14-1)d

⇢280= 7(2a+13d)

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

Now, Substract Equation 1 from Equation 2

=>40-20= 2a+13d-(2a+3d)

=>20= 2a+13d-2a-3d

=>20= 10d

=>d= 20/10=2

hence, common difference (d) is 2

substitute this value into equation 1

=>20= 2a+3(2)

=>20= 2a+6

=>2a= 14

=>a= 14/2=7

a= 7

Now,we have to find sum of first n terms

⇢sₙ=n/2(2a+(n-1)d

⇢sₙ= n/2(2(7)+(n-1)2

⇢sₙ= n/2(14+2n-2)

⇢sₙ=n/2(12+2n)

⇢sₙ= n/2 [ 2(6+n)]

⇢sₙ= n(n+6)

sₙ= n²+6n

Hence,sum of its n terms is n²+6n