The 1st term of an infinite GP exceed s the 2nd term. Buy 2 and its sum is 50 find the series
Please check what you wrote
what is a and what is r in
a + a r + a r^2 + a r^3 ......
If I can inject a little sense into that, how about
ar = a+2
a/(1-r) = 50
That has no real solution, but if you want a sum of -50, you can try
a = -10
r = 4/5
-10, -8, -32/5, -128/25, ...
To find the series, we need to determine the common ratio and the first term of the geometric progression (GP).
Let's denote the first term as 'a' and the common ratio as 'r'.
According to the problem, the 1st term exceeds the 2nd term. So, we have:
a > ar
Dividing both sides of the inequality by 'a', we get:
1 > r
This tells us that the common ratio must be less than 1.
Now, let's write down the two equations based on the given information:
2 terms: a + ar = 50 -- equation (1)
1st term exceeds 2nd term: a > ar -- equation (2)
From equation (1), we can factor out 'a':
a(1 + r) = 50
From equation (2), we can divide both sides by 'a':
1 > r
We also know that the sum of an infinite geometric series is given by:
Sum = a / (1 - r)
Since we are given the sum of the first two terms, we can substitute the values into the equation:
50 = a / (1 - r)
Simplifying this equation, we get:
a = 50(1 - r)
Substituting this value of 'a' in equation (1), we have:
50(1 - r)(1 + r) = 50
Expanding and simplifying, we get:
50 - 50r^2 = 50
Rearranging the terms, we have:
50r^2 = 0
Dividing both sides by 50, we get:
r^2 = 0
Taking the square root of both sides, we find:
r = 0
Since the common ratio 'r' is zero, the series becomes an arithmetic progression rather than a geometric progression.
Therefore, the series is simply: 0, 0, 0, 0, ...
Please note that the sum of this series is zero since all terms are zero.