Answer:
f(5) = 20
f(6) = 40
Step-by-step explanation:
The [tex]$ n^{th} $[/tex] term of a GP is given by:
[tex]$ a_n = ar^{n - 1} $[/tex]
where 'a' is the first term and
'r' is the common difference.
It is given that f(3) = 5 and f(4) = 10
[tex]$ \implies ar^2 = 5 $[/tex]
and [tex]$ ar^3 = 10 $[/tex].
Dividing them we get:
[tex]$ \frac{ar^3}{ar^2} = \frac{10}{5} $[/tex]
[tex]$ \implies r = 2 $[/tex]
i.e, Common difference, r = 2
Now, [tex]$ a_5 = f(5) = ar^4 = ar^3. r = 10(2) = $[/tex] 20
Similarly, [tex]$ a_6 = f(6) = ar^5 = ar^4.r = 20(2) = $[/tex] 40
Hence, the answer.
