> #mathematics , #algebra , #geometric-progression > How many terms in GP 4,3.6,3.24... are needed so that the sums exceeds 53?
Mar 26

# How many terms in GP 4,3.6,3.24... are needed so that the sums exceeds 53?

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Mar 26
To find the number of terms needed in the geometric progression (GP) to exceed a certain sum, we can use the formula for the sum of the first n terms of a GP:

S_n = a * (1 - r^n) / (1 - r)

where:
S_n = sum of the first n terms
a = first term of the GP
r = common ratio of the GP
n = number of terms

In this case, the first term (a) is 4 and the common ratio (r) is 3.6 / 4 = 0.9. We want to find the number of terms (n) needed so that the sum exceeds 53.

Plugging in the values, we get:

53 < 4 * (1 - 0.9^n) / (1 - 0.9)

Simplifying:

53 < 4 * (1 - 0.9^n) / 0.1

53 * 0.1 < 1 - 0.9^n

5.3 < 1 - 0.9^n

-4.3 < -0.9^n

4.3 > 0.9^n

Taking the natural logarithm of both sides:

ln(4.3) > n * ln(0.9)

n < ln(4.3) / ln(0.9)

n < 6.25

Therefore, we need at least 7 terms in the GP for the sum to exceed 53.
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