Question

First term= 2 and 10th term=1024 find the common ratio.

Answer 1

Answer:

2

Step-by-step explanation:

Given:

a = 2

a(10) = 1024

n = 10

We know that the nth term of a G.P. is given by

$a{r}^{n - 1}$

ATQ, the 10th term of the G.P. is

$(2){r}^{10 - 1} = 1024$

We divde both sides by 2

${r}^{9} = 512$

We bring the RHS to the same power as of LHS

${r}^{9} = {2}^{9}$

We get the ninth root of both sides

$r = 2$

Hence the common ratio is 2

Answer 2

Answer:

r =2

Step-by-step explanation:

tn= last term

a= first term

r = ratio

n= number of term

geometric progression last term formula

tn = a×r^(n-1)

1024=2×r^(10-1)

1024/2= r^9

512=r^9

2^9 = r^9

r=2