Question

calculate the number of the term of the given geometric progression: 128,64√2,64,..is √2?​

Answer 1

Answer:

14

Step-by-step explanation:

Given :

128, 64√2, 64, ....... √2 are in GP

First term ( a ) = 128

Common ratio ( r ) = 64√2 / 128 = √2 / 2 = 1 / √2

Let nth term of GP a(n) = √2

Using nth term of GP formula

a(n) = ar^( n - 1 )

⇒ √2 = 128 × ( 1 / √2 )^( n - 1 )

⇒ √2 / 128 = ( 1 / √2 )^( n - 1 )

⇒ 2 / 128√2 = ( 1 / √2 )^( n - 1 )

⇒ 1 / 64√2 = ( 1 / √2 )^( n - 1 )

⇒ 1 / 2^6 × √2 = ( 1 / √2 )^( n - 1 )

⇒ 1 / ( √2 )^12 × √2 = ( 1 / √2 )^( n - 1 )

⇒ 1 / ( √2 )^13 = ( 1 / √2 )^( n - 1 )

⇒ ( 1 / √2 )^13 = ( 1 / √2 )^( n - 1 )

Since bases are equal we can equate exponents

⇒ 13 = n - 1

⇒ n = 14

Therefore there are 14 terms in GP.