Question

find the 9th and 15th term of the series 128,64,32,16
​

Answer 1

Answer:

t₉ = $\frac{1}{2}$

t₁₅ = $\frac{1}{128}$      

Step-by-step explanation:

Given GP , 128,64,32,16

=> a = 128

    r = t₂ / t₁ = 64/128 = 1/2

n'th term of a G.P is given by tₙ = $a*r^{n-1 }$

=> t₉ = 128* $(\frac{1}{2}) ^{9-1}$  = 128*$(\frac{1}{2}) ^{8}$  = $\frac{2^7}{2^8}$ = $\frac{1}{2}$

=> t₁₅    = 128*$(\frac{1}{2}) ^{15-1}$ =   128*$(\frac{1}{2}) ^{14}$ =   $\frac{2^7}{2^14}$ =  $\frac{1}{2^7}$ = $\frac{1}{128}$