The first term of a geometric progression is twice its common ratio. Find the sum…

Explanation

Let the common ratio be r so that the first term is 2r.

Sum, s = \(\frac{a}{1-r}\)
ie. 8 = \(\frac{2r}{1-r}\)

8(1-r) = 2r,

8 - 8r = 2r

8 = 2r + 8r

8 = 10r

r = \(\frac{4}{5}\).

where common ratio (r) = \(\frac{second term(n_2)}{first term(a)}\),

r = \(\frac{n_2}{2r}\)

r = \(\frac{4}{5}\) and a = 2r or \(\frac{8}{5}\)

\(\frac{4}{5}\) * \(\frac{8}{5}\) = n\(_2\)

\(\frac{32}{25}\) = n\(_2\)

The sum of the first two terms = a + n\(_2\)

= \(\frac{8}{5}\) + \(\frac{32}{25}\)

= \(\frac{40 + 32}{25}\)

= \(\frac{72}{25}\)