Question

If the nth term of a geometrical progression is P then what is the multiplication of first (2n-1)numbers​

Answer 1

Hey friend here is the solution for this question.....

lets assume first term is "a" and "r" is the common ratio.

so

$p = a \times {r}^{n - 1}$

Multiplication of first (2n-1) numbers will be....

$= a \times a {r}^{1} \times a \times {r}^{2} .....a \times {r}^{2n - 2}$

$= {a}^{2n - 1} \times {r}^{1 + 2 + 3 ......(2n - 2)}$

sum of 1+2+3....(2n-2)=

$= \frac{2n - 2}{2} \times (1 + (2n - 2)) \\ = (n - 1) \times (2n - 1)$

now multiplication continues as...

$= {a}^{2n - 1} \times {r}^{1 + 2 + 3 ......(2n - 2)}$

$= {a}^{2n - 1} \times {r}^{(n - 1)(2n - 1)} \\ = {(a \times {r}^{n - 1}) }^{2n - 1}$

$= {p}^{2n - 1}$

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