Question

Find the number of terms in a GP given that its first and last terms are 5.5K and 243/256k respectively and that it's common ration is 3/4

Answer 1

Answer:

I am assuming k means 1000 .

Let number of terms n .

$a _n = a {r}^{n - 1} \\ \\ \frac{243}{256} \times 1000 = 5.5 \times1000 \times ( \frac{3}{4} ) {}^{n - 1} \\ \\ ( \frac{ {3}^{5} }{ {4}^{4} } ) = \frac{11}{2} \times ( \frac{3}{4} ) {}^{n - 1} \\ ( \frac{3}{4} ) {}^{n - 1 - 4} = \frac{6}{11}$

So there is no such integer