Question

If the sum of three consecutive terms of an A.P. is $27$ their product is $405$ then find the increasing terms of an A. P.

Answer 1

Let a-r, a, a+r be three terms of A.P

where a is first term 4 r is common difference according to question.

a-r+a+a+r=27

$3a=27\Rightarrow [a=9]$

and $(a-r)(a+r)a=405$

$a(a^2-r^2)=405$

$a^3-a{r}^2=405$

$(a{)}^3-405=a{r}^2$

$729-4005=a{r}^2$

$r^2=\frac{324}{9}=36$

$[r=\pm 6]$

Consider AP is increasing so r=+6

Now first term of AP =a=9 & common difference r=6.

AP= 9+15+21+27+........+6n+3.