Five-distinct-2-digit-numbers-are-in-a-geometric-progression-Find-the-middle-term-

Question Number 20726 by Tinkutara last updated on 01/Sep/17

Five distinct 2 - digit numbers are in a

geometric progression . Find the middle

term .

Answered by dioph last updated on 02/Sep/17

Let the GP be {a, q a, q^{2} a, q^{3} a, q^{4} a}

As we are looking for the middle term

the order is not relevant so we can

safely assume q > 1 .

We know that a ⩾ 10 and q^{4} a < 100,

so q < 2 .

1 < q < 2 \Rightarrow q = \frac{m + n}{n} (m < n both natural numbers)

Because every term is natural, we

must have :

n ∣ a, n^{2} ∣ a, n^{3} ∣ a and n^{4} ∣ a

a < 100 \Rightarrow n \in {2, 3}

* n = 2 \Rightarrow a \in {16, 32, 64} and q = \frac{3}{2},

but {(\frac{3}{2})}^{3} \times 32 = 108 > 100 and

{(\frac{3}{2})}^{2} \times 64 = 144 > 100, hence

in this case a = 16 .

* n = 3 \Rightarrow a = 81 and q ⩾ \frac{4}{3}, but

(\frac{4}{3}) \times 81 = 108 > 100 .

Therefore, n = 2, a = 16 and q = \frac{3}{2},

so GP is {16, 24, 36, 54, 81} and the

answer is 36

Commented by Tinkutara last updated on 02/Sep/17

Thank you very much Sir!