Question

In an a.p, S5 is 55 and S10 is 235 find the sum of its first 20 terms

Answer 1

Answer:

ANSWER

Let the first term is a and the common difference is d

By using S

n

=

2

n

[2a+(n−1)d] we have,

S

5

=

2

5

[2a+(5−1)d]=

2

5

[2a+4d]

S

7

=

2

7

[2a+(7−1)d]=

2

7

[2a+6d]

Given: S

7

+S

5

=167

∴

2

5

[2a+4d]+

2

7

[2a+6d]=167

⇒10a+20d+14a+42d=334

⇒24a+62d=334       ...(1)

S

10

=

2

10

[2a+(10−1)d]=5(2a+9d)

Given: S

10

=235

So 5(2a+9d)=235

⇒2a+9d=47       ...(2)

Multiply equation (2) by 12, we get

24a+108d=564....(3)

Subtracting equation (3) from (1), we get

−46d=−230

 ∴d=5

Substing the value of d=5 in equation (1) we get

2a+9(5)=47 or 2a=2

 ∴a=1

Then A.P is 1,6,11,16,21,⋯