given-the-sum-of-the-first-n-terms-of-an-AP-is-x-2-the-sum-of-the-first-2n-terms-of-the-same-AP-is-x-2-x-show-that-the-sum-of-the-first-4n-terms-is-4x-2-8x-4-

Question Number 34734 by mondodotto@gmail.com last updated on 10/May/18

given the sum of the first n terms

of an AP is x^{2} the sum of

the first 2 n terms of the same AP is x^{2} + x

show that the sum of the first 4 n terms is

4 x^{2} - 8 x + 4

Commented by abdo mathsup 649 cc last updated on 11/May/18

i f t h e f i r s t t e r m i s u_{1} w e h a v e

\sum_{k = 1}^{n} u_{k} = x^{2} \Rightarrow \frac{n}{2} (u_{1} + u_{n}) = x^{2} \Rightarrow

u_{1} + u_{n} = \frac{2 x^{2}}{n} \Rightarrow u_{1} + (n - 1) r = \frac{2 x^{2}}{n} \Rightarrow

(n - 1) r = \frac{2 x^{2}}{n} - u_{1} \Rightarrow r = \frac{2 x^{2} - {n u}_{1}}{n - 1}

r d e p e n d s o n n r i s n o t c o n s t a n t t h e Q . c o n t a i n

a e r r o r \dots .

Answered by Rasheed.Sindhi last updated on 10/May/18

a : first term, d : common difference

Sum of n terms :

\frac{n}{2} [2 a + \overset{―}{n - 1} d] = x^{2}

2 a n + n (n - 1) d = 2 x^{2} \dots . I

Sum of 2 n terms :

\frac{2 n}{2} [2 a + \overset{―}{2 n - 1} d] = x^{2} + x

2 a n + n (2 n - 1) d] = x^{2} + x \dots II

I - II : [n (n - 1) - n (2 n - 1)] d = x^{2} - x

n (n - 1 - 2 n + 1) d = x^{2} - x

n (- n) d = x^{2} - x

- n^{2} d = x^{2} - x

d = \frac{x - x^{2}}{n^{2}}

\frac{n}{2} [2 a + \overset{―}{n - 1} d] = x^{2}

\Rightarrow n [2 a + (n - 1) (\frac{x - x^{2}}{n^{2}})] = 2 x^{2}

2 a n = 2 x^{2} - \frac{n - 1}{n^{2}} (x - x^{2})

a = \frac{x^{2}}{n} - \frac{n - 1}{2 n^{3}} (x - x^{2})

a = \frac{2 n^{2} x^{2} - (n - 1) (x - x^{2})}{2 n^{3}}

Sum of 4 n terms :

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

= 2 n [2 (\frac{2 n^{2} x^{2} - (n - 1) (x - x^{2})}{2 n^{3}}) + (4 n - 1) (\frac{x - x^{2}}{n^{2}})]

= 2 (\frac{2 n^{2} x^{2} - (n - 1) (x - x^{2})}{n^{2}}) + 2 n (4 n - 1) (\frac{x - x^{2}}{n^{2}})

= 2 (\frac{2 n^{2} x^{2} - (n - 1) (x - x^{2}) + 2 n (4 n - 1) (x - x^{2})}{n^{2}})

= \frac{4 n^{2} x^{2} - 2 (n - 1) (x - x^{2}) + 4 n (4 n - 1) (x - x^{2})}{n^{2}}

= \frac{4 n^{2} x^{2} - 2 n x + 2 {n x}^{2} + 2 x - 2 x^{2} + 16 n^{2} x - 16 n^{2} x^{2} - 4 n x + 4 {n x}^{2}}{n^{2}}

The answer is not free of n

Perhsps {there}^{'} s an error \dots

Can you check my solution .

C o n t i n u e

Commented by mondodotto@gmail.com last updated on 10/May/18

please finish it up