# What is the sum : Sum 3*(3/4)^r if r=1 to r=10. the terms are in geometric progression.

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We'll write the sum of the 10 terms of the given geometric progression.

a1 + a2 + .... + a10 = (3/4)^1 + (3/4)^2 + .... + (3/4)^10

We notice that the first term a1 = 3/4 and the common ratio is r = 3/4. Since the common ratio is smaller than 1, we'll apply the following formula:

S10 = a1* [1 - (3/4)^10]/[1- (3/4)]

S10 = (3/4)*[1 - (3/4)^10]/(1/4)

S10 = 3*[1 - (3/4)^10]

We'll multiply the result by 3, both sides:

3*S10 = 3*3*[1 - (3/4)^10]

3*S10 = 9*[1 - (3/4)^10]

**The requested sum of the 10 terms of the geometric progression is:3*S10 = 9*[1 - (3/4)^10].**