If the third term of a GP is 4, then the product of first five terms of the GP is
Let the first term of the GP be ‘a’ and the common ratio be ‘r’.
The terms of the GP are:
1st term: $T_1 = a$
2nd term: $T_2 = ar$
3rd term: $T_3 = ar^2$
4th term: $T_4 = ar^3$
5th term: $T_5 = ar^4$
We are given that the third term of the GP is 4.
$T_3 = ar^2 = 4$.
We need to find the product of the first five terms of the GP.
Product P = $T_1 \times T_2 \times T_3 \times T_4 \times T_5$
P = $a \times (ar) \times (ar^2) \times (ar^3) \times (ar^4)$
Let’s group the ‘a’ terms and the ‘r’ terms:
P = $(a \times a \times a \times a \times a) \times (r^0 \times r^1 \times r^2 \times r^3 \times r^4)$
P = $a^5 \times r^{(0+1+2+3+4)}$
P = $a^5 \times r^{10}$
We can rewrite this expression by grouping $(ar^2)$ terms:
P = $a^5 \times r^{10} = (a^5 \times r^{10/5 \times 5}) = (a^5 \times r^{2 \times 5})$
P = $(a \times r^2)^5$
P = $(ar^2)^5$
We know that $ar^2 = 4$.
Substitute the value of $ar^2$ into the expression for P:
P = $(4)^5 = 4^5$.
– The product of the first n terms of a GP is $a^n r^{n(n-1)/2}$.
– The product of the first 5 terms is $a^5 r^{10}$.
– This can be rewritten as $(ar^2)^5$, which utilizes the given third term.