Question

if r=1,then the sum of first n terms of GP is
a) a(r^n-1)/r-1
b) ar^n-1
c) a(1-r^n)/1-r
d) na

Answer 1

GIVEN :–

• r = 1 for a G.P.

TO FIND :–

• Sum of n term of G.P. = ?

SOLUTION :–

• We know that –

• Sum of n term of G.P. –

$\\ \implies\large { \boxed{\bf S_n = \dfrac{a( {r}^{n} - 1)}{(r - 1)}}}$

• Here –

$\\ \: \: { \huge{.}} \: \: \bf a = first \: \: term$

$\\ \: \: { \huge{.}} \: \: \bf r= common \: \: ratio$

▪︎ Now let's put r=1 :–

$\\ \implies \bf S_n = \dfrac{a( {(1)}^{n} - 1)}{(1 - 1)}$

$\\ \implies \bf S_n = \dfrac{a(1 - 1)}{(1 - 1)}$

$\\ \implies \bf S_n = \dfrac{0}{0}$

• It's an undefined form. So , Let's use limit property –

$\\ \implies\bf S_n = \lim_{r \to1}\dfrac{a( {r}^{n} - 1)}{(r - 1)}$

• Use L'HOSPITAL rule –

$\\ \implies\bf S_n = \lim_{r \to1}\dfrac{a( n{r}^{n - 1} )}{(1)}$

• Now Apply limit –

$\\ \implies\bf S_n =\dfrac{a \{n{(1)}^{n - 1} \}}{(1)}$

$\\ \implies\bf S_n =a \{n{(1)}^{n - 1} \}$

$\\ \implies\bf S_n =a \{n\}$

$\\ \implies \large{ \boxed{\bf S_n =na}}$

• Hence , Option (d) is correct .