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#geometric-progression
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How many terms in GP 4,3.6,3.24.... are needed so that the sum exceeds 35?

Mar 26

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Mar 26

To find the number of terms needed in the geometric progression (GP) so that the sum exceeds 35, we can use the formula for the sum of the first n terms of a geometric progression:

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 35:

4 * (1 - 0.9^n) / (1 - 0.9) > 35

4 * (1 - 0.9^n) / 0.1 > 35

40 * (1 - 0.9^n) > 35

1 - 0.9^n > 35 / 40

1 - 0.9^n > 0.875

0.9^n < 0.125

n * log(0.9) < log(0.125)

n < log(0.125) / log(0.9)

n < 4.292

Therefore, we need at least 5 terms in the geometric progression so that the sum exceeds 35.

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 35:

4 * (1 - 0.9^n) / (1 - 0.9) > 35

4 * (1 - 0.9^n) / 0.1 > 35

40 * (1 - 0.9^n) > 35

1 - 0.9^n > 35 / 40

1 - 0.9^n > 0.875

0.9^n < 0.125

n * log(0.9) < log(0.125)

n < log(0.125) / log(0.9)

n < 4.292

Therefore, we need at least 5 terms in the geometric progression so that the sum exceeds 35.