Given:
[tex]\sum_{k=1}^{6}\dfrac{1}{4}(2)^{k-1}[/tex]
To find:
The values of r and [tex]a_1[/tex].
Solution:
We have,
[tex]\sum_{k=1}^{6}\dfrac{1}{4}(2)^{k-1}[/tex]
Here,
[tex]a_k=\dfrac{1}{4}(2)^{k-1}[/tex] ...(i)
We know that, kth term of a GP is
[tex]a_k=a_1(r)^{k-1}[/tex] ...(ii)
From (i) and (ii), we get
[tex]a_1=\dfrac{1}{4}[/tex] and [tex]r=2[/tex]
Therefore, the correct option is C.
