# the sum to infinity of a gp is four times the first term. find the common ratio

## To find the common ratio (r) of a geometric progression (GP) when the sum to infinity (S∞) is four times the first term (a), we can use the formula for the sum of an infinite geometric series:

S∞ = a / (1 - r)

Given that S∞ = 4a, we can substitute this into the formula:

4a = a / (1 - r)

To solve for the common ratio (r), we can cross-multiply:

4a(1 - r) = a

Now we can simplify and solve for r:

4a - 4ar = a

Multiplying through by -1 to eliminate the negative term:

-4ar + 4a = -a

Factoring out a from both terms:

a(-4r + 4) = -a

Dividing both sides by -a:

-4r + 4 = -1

Subtracting 4 from both sides:

-4r = -5

Dividing both sides by -4:

r = 5/4

Therefore, the common ratio of the geometric progression is 5/4.

## To find the common ratio (r) of a geometric progression (GP) in this case, where the sum to infinity is four times the first term, we can use the formula for the sum to infinity of a GP:

S ∞ = a / (1 - r), where S ∞ is the sum to infinity, a is the first term, and r is the common ratio.

In this case, we are given that S ∞ = 4a. Substituting this into the formula, we have:

4a = a / (1 - r).

To solve for r, we can cross-multiply:

4a × (1 - r) = a.

Simplifying the equation:

4a - 4ar = a.

Rearranging the equation:

4ar + a = 4a.

Factoring out 'a' on the left-hand side:

a(4r + 1) = 4a.

Now, we can divide both sides of the equation by 'a':

4r + 1 = 4.

Subtracting 1 from both sides:

4r = 3.

Finally, dividing both sides by 4:

r = 3/4.

The common ratio (r) of the geometric progression is 3/4.

## sum to infinity = a/(1-r)

but a/(1-r) = 4a
divide both sides by a

1/(1-r) = 4

continue