Question

There are 5 numbers in geometric progression, sum of first three terms is 14 and last two terms is 48 find the numbers

Answer 1

Let the terms be $a,ar,ar^2,ar^3,ar^4$
$a+ar+ar^2=a( \frac{r^3-1}{r-1} )=14$
$ar^3+ar^4=48$
$a+ar+ar^2+ar^3+ar^4= a(\frac{r^5-1}{r-1})=14+48= 62$
$\frac{r^5-1}{r^3-1} = \frac{62}{14}$
$14r^5-62r^3+48=0$
u get real r=2  (just sub. 2 and saw)
which satisfies and further a=2
therefore the terms are 2,4,8,16,32