Question

Find five numbers in A.P.whose sum is 25 by 2 and ratio of first to last terms is 2:3.

Answer 1

let the first term be x and common difference be a
then the 5 terms are x,x+a,x+2a,x+3a,x+4a

x:x+4a=2:3
x/(x+4a)=2/3
3x=2x+8a
x=8a

Sum of 5 terms of A.P= (x+x+4a)*5/2=25/2

(2x+4a)=5
x+2a=5/2
8a+2a=5/2
10a=5/2
a=1/4
x=8/4=2

The terms are 2, 2+1/4 , 2+1/2, 2+3/4,2+1
=2, 9/4,5/2,11/4,3

Answer 2

Answer:

The five terms are $2, \frac{9}{4}, \frac{5}{2}, \frac{11}{4}, 3$.

Step-by-step explanation:

Let the first five terms are

$a,a+d,a+2d,a+3d,a+4d$

It is given that ratio of first and last term is 2:3. It means

$\frac{a}{a+4d}=\frac{2}{3}$

$3a=2a+8d$

$3a-2a=8d$

$a=8d$

The sum of these five terms is $\frac{25}{2}$.

$a+a+d+a+2d+a+3d+a+4d=\frac{25}{2}$

$5a+10d=\frac{25}{2}$

$5(8d)+10d=\frac{25}{2}$

$40d+10d=\frac{25}{2}$

$50d=\frac{25}{2}$

$100d=25$

$d=\frac{25}{100}=\frac{1}{4}$

The value of d is 1/4. So the value of a is

$8d=8\times \frac{1}{4}=2$

Since the value of a is 2 and the value of d is 1/4, therefore the five terms of AP are

$2, 2+(\frac{1}{4}), 2+2(\frac{1}{4}), 2+3(\frac{1}{4}), 2+4(\frac{1}{4})$

$2, \frac{9}{4}, \frac{5}{2}, \frac{11}{4}, 3$

Thus the five terms are $2, \frac{9}{4}, \frac{5}{2}, \frac{11}{4}, 3$.