Question

9) The third, sixth and the last term of a G.P. are 6, 48 and 3072. Find its first term and the
common ratio​

Answer 1

Answer:

a=3/2,r=2

Step-by-step explanation:

General term =(ar)power(n-1)so by equating 3rd term and 6th term we get a and r

Answer 2

The first term is $a=\dfrac{3}{2}$ and common ratio is r=2.

Step-by-step explanation:

It is given that

3rd term : $a_3=6$

6th term : $a_6=48$

nth term of a GP is

$a_n=ar^{n-1}$

Substitute n=3 in the above formula.

$a_3=ar^{3-1}$

$6=ar^{2}$            .... (1)

Substitute n=6 in the above formula.

$a_6=ar^{6-1}$

$48=ar^{5}$           .... (2)

Divide equation (2) be equation (1).

$\dfrac{48}{6}=\dfrac{ar^{5}}{ar^2}$

$8=r^3$

$2^3=r^3$

$r=2$

Substitute r=2 in equation (1).

$6=a(2)^{2}$

$6=4a$

$\dfrac{6}{4}=a$

$\dfrac{3}{2}=a$

Therefore, the first term is $a=\dfrac{3}{2}$ and common ratio is r=2.

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