Question

The sum of a GP with common ratio 3 is 364 and last term is 243 , then the numbers of terms is

• A. $6$
• B. $5$
• C. $4$
• D. $10$

Answer 1

Answer: (A)

$6$

$nth$ terms=$a{r}^{(n-1)}=a\times 3^{(n-1)}$

$=>a\times 3^{(n-1)}=243$ [now multiply both side by 3]

$a\times 3^{(n-1)}\times 3=243\times 3=729$

$=>a\times 3^{\left(n\right)}=729$

sum=$a\frac{1-r^n}{1-r}=a\left[\frac{1-729}{\frac{1}{3}}\right]$

$=>\frac{-728}{-2}=364a$

$=>364=364a$

$=>a=1$

Since,

$=>a\times 3^n=729$ and $a=1$

$=>3^n=729$

$=>\log 3^n=n\times \log 3=\log 729$

$=>\frac{\log 729}{\log 3}$

$=>\frac{\log 3^6}{\log 3}$

$=>\frac{6\log 3}{\log 3}$

$=>n=6$