Question

Find five numbers in an Arithmetic Progression whose sum is 12 and a half, and the ratio of first term to last term is 2:3

Answer 1

Using the formula for an Arithmetic Progression:
$$
\large{
\ a_n = a_1 + (n - 1)d\\
\cfrac{a_5}{a_1}=\cfrac{a_1+4d}{a_1}=\cfrac{3}{2}\\
}
$$
http://en.wikipedia.org/wiki/Arithmetic_series

And for their sum:
$$
\large{
\frac{5(a_1 + a_5)}{2}=\frac{5(a_1 + a_1+4d)}{2}12=\cfrac{1}{2}
}
$$

You now have 2 equations with 2 unknowns. Solvable.

Answer 2

Let the five numbers in AP be:

a-2d, a-d, a, a+d, a+2d (where a is the first term and d is the common difference)
Now according to the question, their sum
i.e. a-2d+a-d+ a+ a+d+ a+2d=12
i.e. 5a = 12
i.e. a= 12/5
i.e. a= 2.4

Now since, the ratio of first term to last term is 2:3
therefore
\[\frac{a-2d}{a+2d} = \frac{2}{3} \rightarrow \frac{2.4-2d}{2.4+2d} = \frac{2}{3}\]
i.e. 7.2-6d= 4.8 +4d
i.e. 7.2-4.8 =6d +4d
i.e. 2.4 =10d
i.e. d= 2.4/ 10
i.e. d= 0.24

Hence:
first term =a-2d = 2.4 - 2*0.24 = 2.4- 0.48 =1.92

second term =a-d = 2.4 - 0.24 =2.16

Third term = a = 2.4

fourth term = a+d = 2.4+0.24 =2.64

fifth term = a+2d = 2.4 + 2*0.24 = 2.4+ 0.48 =2.88

Hence the five numbers in AP are 1.92, 2.16, 2.4, 2.64, 2.88

@hackpert

Answer 3

Thanks a lot for the help, @dpasingh.