c) The sum to infinity of a GP is twice the sum of the first two terms. Find possible values of the common ratio
Let's denote the first term of the geometric progression (GP) as 'a' and the common ratio as 'r'.
The sum of the first two terms is given by:
S2 = a + ar = a(1 + r)
The sum to infinity of the GP is calculated using the formula:
S∞ = a / (1 - r), where |r| < 1.
Now, according to the given condition, we can write:
S∞ = 2 * S2
Substituting the expressions for S∞ and S2, we get:
a / (1 - r) = 2 * (a(1 + r))
Simplifying this equation, we have:
a / (1 - r) = 2a(1 + r)
Dividing both sides by 'a' (assuming a is non-zero), we get:
1 / (1 - r) = 2(1 + r)
Now, let's solve for 'r':
1 - r = 2(1 + r)
Expanding the right side, we have:
1 - r = 2 + 2r
Moving all 'r' terms to the left side, we get:
3r = 1 - 2
Simplifying further, we have:
3r = -1
Finally, we can determine the value of 'r' by dividing both sides by 3:
r = -1/3
Therefore, the possible value of the common ratio is -1/3.
To find the possible values of the common ratio, we can use the formula for the sum to infinity of a geometric progression (GP). Let's denote the first term as 'a' and the common ratio as 'r'.
The sum to infinity of a GP is given by the formula:
S = a / (1 - r)
According to the problem, the sum to infinity is twice the sum of the first two terms. This can be written as:
2a = a + ar
Simplifying the equation, we have:
2 = 1 + r
Rearranging the equation to solve for 'r', we get:
r = 2 - 1
r = 1
Therefore, the possible value of the common ratio 'r' is 1.
S = a + a r
so 2S = 2 a + 2 a r
and
2 S = a / (1+r)
so
2 a + 2 a r = a /(1+r)
4 = 1/(1+r)
4 + 4 r = 1
r = -3/4