Question

Which term of the gp √2,1/√2,1/2√2,1/4√2,............is 1/512√2

Answer 1

Answer:

Step-by-step explanation:

√2 , 1/2√2 , 1/4√2 ......

Tₙ = 1/512√2

We know that Tₙ = a . rⁿ⁻¹

a = √2, r = 1/2

=> √2 * (1/2)ⁿ⁻¹ = 1/512√2

=> (1/2)ⁿ⁻¹ = 1/512√2 * 1/√2

=> (1/2)ⁿ⁻¹ = 1/512 * 1/2

=> (1/2)ⁿ⁻¹ = 1/1024

=> (1/2)ⁿ⁻¹ = (1/2)¹⁰

//when bases are same powers are equal

=> n - 1 = 10

=> n = 11.

Hence it is the 11th term which is equal to 1/512√2

Answer 2

Answer:

11th

Step-by-step explanation:

general term of a gp is $ar^{n-1}$

so a= $\sqrt{2}$ , and $a_{n}$ = $512\sqrt{2}$

r = $\frac{1}{\sqrt{2} } /\sqrt{2} = 1/2$

therefore

$\sqrt{2} * \frac{1}{2^{n-1} } = \frac{1}{512*\sqrt{2} }$

Solving which we get,

n-1 = 10

therefore n =11