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


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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
	$Let the GP be {a,qa,q^2 a,q^3 a,q^4 a} As we are looking for the middle term the order is not relevant so we can safely assume q > 1. We know that a ≥ 10 and q^4 a < 100, so q < 2. 1 < q < 2 ⇒ q = ((m+n)/n) (m<n both natural numbers) Because every term is natural, we must have: n ∣ a, n^2 ∣ a, n^3 ∣ a and n^4 ∣ a a < 100 ⇒ n ∈ {2, 3} ∗ n = 2 ⇒ a ∈ {16, 32, 64} and q = (3/2), but ((3/2))^3 ×32 = 108 > 100 and ((3/2))^2 ×64 = 144 > 100, hence in this case a = 16. ∗ n = 3 ⇒ a = 81 and q ≥ (4/3), but ((4/3))×81 = 108 > 100. Therefore, n = 2, a = 16 and q = (3/2), so GP is {16, 24, 36, 54, 81} and the answer is 36$
	$Let the GP be {a, q a, q^{2} a, q^{3} a, q^{4} a}$ $As we are looking for the middle term$ $the order is not relevant so we can$ $safely assume q > 1 .$ $We know that a ⩾ 10 and q^{4} a < 100,$ $so q < 2 .$ $1 < q < 2 \Rightarrow q = \frac{m + n}{n} (m < n both natural numbers)$ $Because every term is natural, we$ $must have :$ $n ∣ a, n^{2} ∣ a, n^{3} ∣ a and n^{4} ∣ a$ $a < 100 \Rightarrow n \in {2, 3}$ $* n = 2 \Rightarrow a \in {16, 32, 64} and q = \frac{3}{2},$ $but {(\frac{3}{2})}^{3} \times 32 = 108 > 100 and$ ${(\frac{3}{2})}^{2} \times 64 = 144 > 100, hence$ $in this case a = 16 .$ $* n = 3 \Rightarrow a = 81 and q ⩾ \frac{4}{3}, but$ $(\frac{4}{3}) \times 81 = 108 > 100 .$ $Therefore, n = 2, a = 16 and q = \frac{3}{2},$ $so GP is {16, 24, 36, 54, 81} and the$ $answer is 36$
	Commented by Tinkutara last updated on 02/Sep/17

	$Thank you very much Sir!$
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