Question

which term of GP,2,2√2,4 ...... is 128​

Answer 1

The $13^{th}$ term of GP is 128.

• Given GP is 2,2$\sqrt{2}$,4,....
• As the given sequence is in GP we find the common ratio of the GP.
• Here first term(a) =2
• Common Ratio (r) = $\frac{t_2}{t_1}$

r = $\frac{second \ term}{first \ term}$

r = $\frac{2\sqrt{2} }{2}$

r = $\sqrt{2}$

• Common ratio obtained is $\sqrt{2}$ .
• Now we have to find the which term is equal to 128.
• We apply the formula of $n^{th}$ term of GP.

$t_n = ar^{n-1}$

• Here $t_n$ = 128 , a=2 and r = $\sqrt{2}$ .

• Substituting the values in the equation, we get

128 = 2 ${(\sqrt{2})}^{n-1}$

64 = ${(\sqrt{2})}^{n-1}$

• We write 64 in powers of ($\sqrt{2}$)

${2}^{6}$ = ${(\sqrt{2})}^{n-1}$

${({(\sqrt{2})}^{2})}^{6}$ = ${(\sqrt{2})}^{n-1}$

• Equating both sides, we get

12 = n -1

n = 13

• The $13^{th}$ term of GP will be 128.