Identify the missing number in the given sequence 32, 80, 200,___, 1250, ____.

Hint: If you observe the terms in the sequence given in the question, you will notice that the terms are forming an increasing G.P. with the first term as 32 and common ratio equal to $\dfrac{5}{2}$ . Now you are asked to find the fourth and sixth term of the G.P., which you can easily find using the formula ${{T}_{n}}=a{{r}^{n-1}}$ .

Complete step-by-step solution -
Before starting with the solution, let us discuss what an G.P. . G.P. stands for Geometric progression and is defined as a sequence of numbers for which the ratio of two consecutive terms is constant. The general term of a geometric progression is denoted by ${{T}_{n}}$ , and sum till n terms is denoted by ${{S}_{n}}$ .
${{T}_{n}}=a{{r}^{n-1}}$
${{S}_{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}$
Now let us move to the sequence given in the above question. If we take the ration of the consecutive terms, we get
$\dfrac{{{T}_{2}}}{{{T}_{1}}}=\dfrac{80}{32}=\dfrac{5}{2}$
$\dfrac{{{T}_{3}}}{{{T}_{2}}}=\dfrac{200}{80}=\dfrac{5}{2}$
So, the sequence is a G.P. whose common ration is $\dfrac{5}{2}$ . Also, the first term is 32.
Now the missing terms are ${{T}_{4}}\text{ and }{{T}_{6}}$ .So, if we use the formula of general term of a G.P, we get
${{T}_{4}}=32\times {{\left( \dfrac{5}{2} \right)}^{4-1}}=32\times \dfrac{125}{8}=500$
${{T}_{6}}=32\times {{\left( \dfrac{5}{2} \right)}^{6-1}}=32\times \dfrac{3125}{32}=3125$
So, the complete sequence is 32, 80, 200, 500, 1250, 3125.

Note: It is not always necessary that by looking at the sequence you can predict whether it is an G.P. or not, you might need to find the ratios of all the consecutive terms to conclude that the sequence is a G.P. It is also suggested that whenever you come across such a sequence try to find the relation between the consecutive terms, may not be an A.P. or a G.P. but might turn out to be an A.G.P. or some other complex sequence.