If the third term of a GP is 4, then the product of first five terms o

If the third term of a GP is 4, then the product of first five terms of the GP is

4<sup>3</sup>
4<sup>4</sup>
4<sup>5</sup>
4<sup>6</sup>
This question was previously asked in
UPSC CISF-AC-EXE – 2019
A geometric progression (GP) is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.
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 n-th term of a GP is given by $T_n = ar^{n-1}$.
– 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.
In general, for a GP with an odd number of terms (say, $2k+1$ terms), the product of these terms is the $(k+1)$-th term raised to the power of $(2k+1)$. In this case, we have 5 terms (which is $2 \times 2 + 1$), so $k=2$. The product is the $(2+1)=3$rd term raised to the power of 5. Product = $(T_3)^5$.