A geometric progression has first term a ,common ratior and sum to infinity 6. A second geometric progression has first term 2a. common ratio r^2and sum to infinity 7. Find the values
of a and r

a = , r =

Step-by-step explanation:

The sum to infinity of a geometric progression is

; | r | < 1

Thus for first progression

= 6 ( multiply both sides by (1 - r) )

a = 6(1 - r) → (1)

Second progression

= 7 ← multiply both sides by (1 - r² )

2a = 7(1 - r² ) = 7(1 - r)(1 + r) ← difference of squares

2a = 7(1 - r)(1 + r) → (2)

Substitute a = 6(1 - r) into (2)

2(6(1 - r) = 7(1 - r)(1 + r)

12(1 - r) = 7(1 - r)(1 + r) ← divide both sides by (1 - r)

12 = 7(1 + r) = 7 + 7r ( subtract 7 from both sides )

5 = 7r ( divide both sides by 7 )

r =

Substitute this value into (1)

a = 6(1 - ) = 6 × =

the least common denominatore of 7 and 2 is 9

step-by-step explanation:

7+2=9

point a is the center of the circle that passes through points e, f, and g and the center of the circle that passes through points x, y, and z.