Question

A,B,C is form an AP which sum of is 20 find A,B,C in form of no.​

Answer 1

Answer:

The three terms of the AP are $0, \frac{20}{3}\ and\ \frac{40}{3}$.

Step-by-step explanation:

There are three terms in this AP namely, A, B and C.

So n = 3.

The common difference is:

d = B - A = C - B

2B = C - A

The sum of n terms of an AP is:

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

The sum of the terms A, B and C is 20.

$S_{n}=\frac{n}{2}[2a+(n-1)d]\\20=\frac{3}{2}[2A+(3-1)(B-A)]\\40=3[2A+2B-2A]\\40=3\times 2B\\40=6B\\B=\frac{20}{3}$

So, $C-A=2B=\frac{40}{3}...(i)$

And

$A+B+C=20\\C+A=20-\frac{20}{3}\\C+A=\frac{40}{3}...(ii)$

Solve (i) and (ii) as follows:

$C-A=\frac{40}{3}\\C+A=\frac{40}{3}\\\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\2C=\frac{80}{3}\\C=\frac{40}{3}$

Then A is,

$C-A=\frac{40}{3}\\\frac{40}{3}-A=\frac{40}{3}\\A=0$

Thus, the three terms of the AP are $0, \frac{20}{3}\ and\ \frac{40}{3}$.