Question

The third term of a gp is 2. then the product of first 5 terms is

Answer 1

third term =2
$a {r}^{3} = 2..................(1)$

let the terms be
$a \: \: ar {}^{2} \: \: ar {}^{3} \: \: ar {}^{4} \: \: \: ar {}^{5}$
multiply all we get
$= {a}^{5} {r}^{15}$
$= (a {r}^{3} ) {}^{5} \\ from \: (1) \: a {r}^{3} = 2 \: put \: here \: \\ = (2) {}^{5} \\ = 32$
answer is 32
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Answer 2

Given:

The third term of a GP ($a_{3}$) = 2

Let the first term = a and common ratio = r

We have to find the product of first 5 terms of a GP = ?

Solution:

We know that,

The nth term of a GP

$a_{n}$ = $ar^{n-1}$

∴ We know that,

The third term of a GP

$a_{3}$ = $ar^{3-1}$

⇒ $ar^{2}$ = 2          ................. (1)

The product of first 5 terms

= a × ar × $ar^{2}$ × $ar^{3}$ ×  $ar^{4}$

= $a^{5}$ ×  $r^{10}$

= $(ar^{2})^5$

Using equation (1), we get

= $(2)^5$

= 32

∴ The product of first 5 terms = 32

Thus, the product of first 5 terms is equal to 32.