Question

The sum to infinity of a GP is 60 .if the first term write the next 3 term of the GP.

Answer 1

To solve this, we need to use the formula for the sum to infinity of a geometric progression (GP), which is given by:

\[
S = \frac{a}{1 - r}
\]

where \(S\) is the sum to infinity, \(a\) is the first term, and \(r\) is the common ratio of the GP.

Given that the sum to infinity \(S\) is 60, we can write:

\[
\frac{a}{1 - r} = 60
\]

From this, we can express \(a\) in terms of \(r\):

\[
a = 60(1 - r) = 60 - 60r
\]

Now, we need to write the next three terms of the GP. The terms of the GP can be expressed as follows:

1. First term: \(a\)
2. Second term: \(ar\)
3. Third term: \(ar^2\)
4. Fourth term: \(ar^3\)

Thus, the next three terms are:

- Second term: \(ar = (60 - 60r)r = 60r - 60r^2\)
- Third term: \(ar^2 = (60 - 60r)r^2 = 60r^2 - 60r^3\)
- Fourth term: \(ar^3 = (60 - 60r)r^3 = 60r^3 - 60r^4\)

To find specific values for these terms, we need to assume a value for \(r\) where \(|r| < 1\) for the sum to converge. For instance, if we assume \(r = \frac{1}{2}\):

1. Calculate \(a\):
\[
a = 60(1 - \frac{1}{2}) = 60 \times \frac{1}{2} = 30
\]

2. Now calculate the next three terms:
- Second term: \(ar = 30 \times \frac{1}{2} = 15\)
- Third term: \(ar^2 = 30 \times \left(\frac{1}{2}\right)^2 = 30 \times \frac{1}{4} = 7.5\)
- Fourth term: \(ar^3 = 30 \times \left(\frac{1}{2}\right)^3 = 30 \times \frac{1}{8} = 3.75\)

Thus, if we take \(r = \frac{1}{2}\), the first term is 30 and the next three terms of the GP are 15, 7.5, and 3.75.

If you have a different value of \(r\) in mind, you can substitute that back into the formulas to find the particular terms.