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Integer GP, 5 Terms All Under 100: Count Them | JEE

JEE Maths question with a full step-by-step solution.

Question
Let a1,a2,a3,a4,a5a_1, a_2, a_3, a_4, a_5 be five terms of a geometric sequence satisfying
0<a1<a2<a3<a4<a5<1000<a_1<a_2<a_3<a_4<a_5<100
 where each term is an integer. Then the number of such geometric progressions is
A66
B77
C88correct
D1010
Solution
Step 1: Let the ratio be nm\dfrac{n}{m} in lowest terms with n>mn>m. For every term to be an integer, a1a_1 must be a multiple of m4m^4; write a1=k m4a_1=k\,m^4. The terms are
km4, km3n, km2n2, kmn3, kn4.k m^4,\ k m^3 n,\ k m^2 n^2,\ k m n^3,\ k n^4.
 Step 2: The largest satisfies kn4<100kn^4<100. If n≥4n\ge 4, then n4≥256>100n^4\ge 256>100 — impossible, so n≤3.n\le 3. Step 3: The cases:
n=3, m=2: 81k<100⇒k=1: 16,24,36,54,81.n=3,\ m=2:\ 81k<100\Rightarrow k=1:\ 16,24,36,54,81.
n=3, m=1: 81k<100⇒k=1: 1,3,9,27,81.n=3,\ m=1:\ 81k<100\Rightarrow k=1:\ 1,3,9,27,81.
n=2, m=1: 16k<100⇒k=1,…,6: six progressions.n=2,\ m=1:\ 16k<100\Rightarrow k=1,\ldots,6:\ \text{six progressions}.
 Step 4: Total =1+1+6=8.=1+1+6=8. Correct answer: (3)
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