Question

What is b1 and r of geometrical sequence if b3 −b1 = 16 and b5 −b3 = 144.

Answer 1

Answer:

b1 = 2 ; r = 3

Step-by-step explanation:

Given that :

if b3 −b1 = 16 and b5 −b3 = 144.

For a geometric series :

Ist term = a

Second term = ar

3rd term = ar^2

4th term = ar^3

5th term = ar^4 ;...

If b3 - b1 = 16;

ar^2 - a = 16

a(r^2 - 1) = 16 - - - (1)

b5 - b3 = 144

ar^4 - ar^2 = 144

ar^2(r^2 - 1) = 144 - - - - (2)

Divide (1) by (2)

a(r^2 - 1) / ar^2(r^2 - 1) = 16 /144

a / ar^2 = 1 / 9

ar^2 = 9a

Substitute for a in ar^2 - a = 16

9a - a = 16

8a = 16

a = 2

From ar^2 - a = 16

2r^2 - 2 = 16

2r^2 = 16 + 2

2r^2 = 18

r^2 = 18 / 2

r^2 = 9

r = √9

r = 3

Hence ;

a = b1 = 2 ; r = 3