Given that the third term of the GP, a3 = 4
Let us assume the GP mentioned in the question be,
\(\frac{A}{R^3},\)With the first term \(\frac{A}{R}\), A, AR, AR2 ......
With the first term \(\frac{A}{R^2}\) and common ratio R.
Now, the third term in the assumed GP is A.
So, A = 4 (given data)
Now,
Product of the first five terms of GP = \(\frac{A}{R^2}\) \(\times\) \(\frac{A}{R}\) \(\times\) A \(\times\) AR \(\times\) AR2 = A5
So, the required product = A5 = 45 = 1024
∴ The product of first five terms of a GP with its third term 4 is 1024.