Have You Tried Pairing Them Up?

Like the title suggested, let's group the numbers in pairs.

In this geometric progression, we know that the first term $a = 1$ and the $100^\text{th}$ term is $ar^{99} = 100$ , where $r$ is the common ratio.

We want ot find the product of the other 98 numbers.

Let $a_2, a_3, \ldots , a_{99}$ denote these numbers. And we want to find the value of

$a_2 \times a_3 \times \cdots \times a_{99}$

By grouping the numbers as such,

$(a_2 \times a_{99} ) \times (a_3 \times a_{98} ) \times (a_4 \times a_{97} ) \times (a_5 \times a_{96} ) \times \cdots \times (a_{50} \times a_{51} )$

Notice that each of the values inside the bracket can be evaluated as $a_k \times a_{101-k}$ , where $k = 2, 3, 4, \ldots , 50$ .

The $k^\text{th}$ term of a geometric progression can be expressed as $a_k = ar^{k-1}$ , so $a_{101-k} = ar^{100-k}$ . Thus $a_k \times a_{101-k} = a^2 r^{99} = a \times (ar^{99}) = 1\times 100 = 100$ .

In other words, all the values inside the brackets as stated above are all equal to 100.

Since there's $50 - 1 = 49$ terms, then we need to multiply 100 by 49 times, or simply put, the product is equal to $100^{49} = (10^2)^{49} = 10^{2\times49} = \boxed{10^{98}} \; .$

Consider the GP as $1,r,r^2,r^3,\dots ,r^{98},r^{99}$ where r is the common ratio and $r^{99}=100$ . Note that if we see the product of the middle 98 terms, according to the rule of exponents, the indices sum from 1 to 98 which equals $\dfrac{98\times 99}{2}$ which is the index of r. But since $r^{99}=100$ , the product simplifies to $100^{98/2}=\boxed{10^{98}}$

Let the common ratio be $r$ . Then $a_1r^{99}=r^{99}=100$ . Now, the product of $a_2$ to $a_{99}$ is as follows.

$\begin{aligned} P&=\prod_{k=2}^{99} a_k \\ & = r \cdot r^2 \cdot r^3 \cdots r^{98} \\ & = r^{1+2+3+...+98} \\ & = r^{\frac {99\cdot 98}2} \\ & = 100^{\frac {98}2} \\ & = \boxed{10^{98}} \end{aligned}$

Nice question.