Question

2 Find the least positive integer n such that 1+6+62 + ... +6" > 5000
the least number of terms for which the sum must be g​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

We know that the sum of a G.P. is a(r-1)^n/r–1 for r>1.

Putting values along with condition:

1(6–1)^n/6–1 > 5000

5^n/5> 5000

5^(n-1) > 5000

Diving both sides by 5,

5^(n-2) > 1000

5^(n-2) > 125*8

5^(n-2) > 5^3*8

5^(n-5) > 8

Taking log both sides,

Log(5^(n-5)) > Log(2^3)

(n-5)Log5> 3Log2

n-5 > 3Log2/Log5

Putting values of Log2 and Log5,

n-5> 3*(0.301)/0.698

n-5> 1.3

n> 6.3

But n is an integer, so,

n = 7