Question

Which term of 0.004+0.02+0.1=... is 12.5?

Answer 1

This is a G.p with a=0.004 & common ratio r= 5

take
$t(n) = 12.5$
$a {r}^{n - 1} = 12.5$
$(0.004) {5}^{n - 1} = 12.5$
${5}^{n - 1} = 12.5 \div 0.004$
${5}^{n - 1} = 12.5 \times 250$
$= 125 \times 25$
$= {5}^{5}$
$n - 1 = 5$
$n = 6$
sixth term of given GP is 12.5

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Answer 2

Answer:

Here is the answer for your question.

Step-by-step explanation:

By using the formula,

Tn = ar^n-1

Given:

a = 0.004

r = $t_{2}$/$t_1$ = (0.02/0.004)

= 5

$T_{n}$ = 12.5

n = ?

So,  $T_{n}$ = ar^n-1

12.5 = (0.004) (5)^n-1

12.5/0.004 = 5^n-1

3000 = 5^n-1

5^5 = 5^n-1

5 = n-1

n = 5 + 1

= 6

∴ 6th term of the progression 0.004, 0.02, 0.1, …. is 12.5.