#### In an AP...

In an AP with common difference not equal to 0, if the sum of the first 3n terms is equal to the sum of the next n terms, what is the ration of the sum of the first 2n terms to that of the second 2n terms?

Anonymous

0

Sangeetha Pulapaka

0

Did you mean to ask how to find the ratio if the sum of the first 2n terms to that of the sum of the next 2n terms?

Given the arithmetic progression whose common difference is nonzero the sum of first 3n terms is equal to the sum of n terms.

So, we have total = 4n terms

Sum of first 3n terms = Sum of n terms

S_{3n} = Sum of n terms

S_{3n} = S_{4n} - S_{3n}

2S_{3n} = S_{4n}

We know the formula for sum of n terms in an A.P

S_{n} = \frac{n}{2} [2a + (n-1)d]

[math] 2 \times \frac{3n}{2}[ 2a +(3n-1) d] [/math] = [math] \frac{4n}{2}[ 2a+(4n-1)d][/math]

Simpifying this we get 2a + nd = d

\Rightarrow 2a = d ( 1-n)

We have to find the ratio if the sum of the first 2n terms to that of the sum of the next 2n terms

\frac{S_{2n}}{S_{'2n}} = \frac{S_{2n}}{S_{4n}-S_{2n}}

\Rightarrow [math]\frac{\frac{2n}{2}[2a+(n-1)d]}{\frac{4n}{2}[4a+(n-1)d]-{\frac{2n}{2}[2a+(2n-1)d]}}[/math]

Simplifying this we get \frac{n}{5}