HOW MANY TERMS OF GP 1,2,4,8..... GIVE A SUM GREATER THAN 500.
To find out how many terms of a geometric progression (GP) are needed to give a sum greater than 500, we need to find the sum of the series until it exceeds 500.
The general term of a geometric progression is given by the formula:
an = a * r^(n-1)
Given the GP 1, 2, 4, 8, ... we can see that the first term (a) is 1, and the common ratio (r) is 2.
The sum of a GP with a finite number of terms can be calculated using the formula:
Sn = a * (r^n - 1) / (r - 1)
We want to find the smallest value of n for which the sum, Sn, exceeds 500.
Now we can solve the equation:
1 * (2^n - 1) / (2 - 1) > 500
Simplifying the equation, we have:
2^n - 1 > 500
To solve this inequality, we can take the logarithm base 2 on both sides:
log2(2^n - 1) > log2(500)
Simplifying further:
n * log2(2) > log2(500) + log2(1)
Since log2(2) equals 1, the inequality becomes:
n > log2(500)
Using a calculator, we find that log2(500) is approximately 8.965784...
Therefore, the smallest integer value of n that satisfies the inequality is 9.
Hence, 9 terms of the geometric progression 1, 2, 4, 8, ... are needed to have a sum greater than 500.