Question

If the 5th and 10th term of a GP are 32 and 1024 respectively.find the 11th term of the GP

Answer 1

GIVEN :–

• 5th term of G.P. = 32

• 10th term of G.P. = 1024

TO FIND :–

• 11th term of G.P. = ?

SOLUTION :–

• We know that nth term of G.P. –

$\\ \implies \large{ \boxed{\bf T_n = a {r}^{n - 1}}}$

• According to the first condition –

$\\ \implies\bf T_5 =32$

$\\ \implies\bf a {r}^{5- 1}=32$

$\\ \implies\bf a {r}^{4}=32 \: \: \: \: \: \: \: \: \: \: - - - eq.(1)$

• According to the second condition –

$\\ \implies\bf T_{10} =1024$

$\\ \implies\bf a {r}^{10-1}=1024$

$\\ \implies\bf a {r}^{9}=1024$

$\\ \implies\bf (a {r}^{4})({r}^{5}) =1024$

• Using eq.(1) –

$\\ \implies\bf (32)({r}^{5}) =1024$

$\\ \implies\bf {r}^{5} = \cancel\dfrac{1024 }{32}$

$\\ \implies\bf {r}^{5} =32$

$\\ \implies\bf {r}^{5} ={2}^{5}$

$\\ \implies \large{ \boxed{\bf r = 2}}$

• Put the value of 'r' in eq.(1) –

$\\ \implies\bf a{(2)}^{4}=32$

$\\ \implies\bf a(16)=32$

$\\ \implies\bf a= \cancel \dfrac{32}{16}$

$\\ \implies \large{ \boxed{\bf a=2}}$

▪︎ 11th term of G.P. –

$\\ \implies\bf T_{11} = (2){(2)}^{11- 1}$

$\\ \implies\bf T_{11} = (2){(2)}^{10}$

$\\ \implies\bf T_{11} = (2)(1024)$

$\\ \implies \large{ \boxed{\bf T_{11} = 2048}}$