Question

nth term of GP 64,32,16 is​

Answer 1

Answer:

Step-by-step explanation:

r=32/64=1/2

a=64

nth term=ar^(n-1)=(64)*(1/2)^(n-1)

Answer 2

$Given \: G.P : 64 , 32 , 16 ,\cdot \cdot \cdot$

$First \term (a) = 64$

$Common \:ratio (r) = \frac{a_{2}}{a_{1}} \\= \frac{32}{64} \\= \frac{1}{2}$

$\boxed { \pink { n^{th} \:term (a_{n}) = a r^{n-1} }}$

$a_{n} = 64 \times \Big( \frac{1}{2}\Big)^{n-1} \\= \frac{64}{2^{n-1} }$

Therefore.,

$\red { n^{th} \:term (a_{n})} \green { = \frac{64}{2^{n-1} }}$

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