hi-sirs-please-how-to-solve-this-thing-below-a-R-x-1-x-2-a-x-2-x-3-2a-x-n-1-x-n-n-1-a-

Question Number 140834 by greg_ed last updated on 13/May/21

hi, sirs!

please, how to solve this thing below (a \in R)

{\begin{cases} x_{1} - x_{2} = a \\ x_{2} - x_{3} = 2 a \\ \dots \dots \dots \dots \dots . . \\ \dots \dots \dots \dots \dots . . \\ \dots \dots \dots \dots \dots . \\ x_{n - 1} - x_{n} = (n - 1) a \\ x_{1} + x_{2} + \dots + x_{n - 1} + x_{n} = n a \end{cases}

Commented by Rasheed.Sindhi last updated on 13/May/21

A d d i n g a l l t h e e q u a t i o n s :

x_{1} + x_{n} = a (1 + 2 + \dots + n)

x_{1} + x_{n} = \frac{n (n + 1) a}{2}

\dots . .

\dots

Commented by prakash jain last updated on 13/May/21

x_{1} = x_{1}

x_{2} = x_{1} - a

x_{3} = x_{2} - 2 a = x_{1} - (a + 2 a)

x_{4} = x_{1} - (a + 2 a + 3 a)

\dots

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

\sum_{i = 1}^{n} x_{i} = {n x}_{1} - \underset{i = 1}{\sum^{n}} i (i - 1) a

= {n x}_{1} - a \underset{i = 1}{\sum^{n}} i^{2} - a \underset{i = 1}{\sum^{n}} i

= {n x}_{1} - \frac{n (n + 1) (2 n + 1) a}{6} - \frac{n (n + 1)}{2} a

= n a (g i v e n)

x_{1} = a + \frac{(n + 1)}{2} a + \frac{(n + 1) (2 n + 1)}{6} a

= \frac{a}{3} (n^{2} + 3 n + 5)

x_{i} = x_{1} - \frac{i (i - 1)}{2} \times a

Commented by Rasheed.Sindhi last updated on 13/May/21

W \in LCOM \in SIR a f t e r l o n g i n t e r v a l!

Commented by Ar Brandon last updated on 13/May/21

Greetings Sir Prakash Jain .