Question

The 10th term of a G.P. is 320 and 6th term is 20. Find the progression.

Answer 1

Let a be the first term and r be the common ratio.
Now, $a_{10}=320\Rightarrow a{r}^9=320\dots \dots \left(1\right)Now,a_6=20\Rightarrow a{r}^5=20\dots \dots \left(2\right)$
Dividing (1) by (2), we get,
$\frac{r^9}{r^5}=\frac{320}{20}\Rightarrow r^4=16\Rightarrow r=2$
Now, from (2), we get,
$a(2{)}^5=20\Rightarrow a\times 32=20\Rightarrow a=\frac{20}{32}\Rightarrow a=\frac{5}{8}$
Now, $a_2=a_r=\frac{5}{8}\times 2=\frac{5}{4}{a}_3=a{r}^2=\frac{5}{8}\times (2{)}^2=\frac{5}{8}\times 4=\frac{5}{2}$
Hence, required GP is:
$\frac{5}{8},\frac{5}{4},\frac{5}{2},\dots \dots$