Question

The missing numbers in the following series would be:
1 : 1, 8 : 4, 9 : 27, 64 : 16, 25 : 125, ? : ?, 49 : 343

• A. 36 : 316
• B. 216 : 36
• C. 316 : 16
• D. 32 : 316

Hint:

Write each number as a square or cube of 1 to 7; the order of square/cube flips between odd and even n.

Answer 1

The Correct Option is B

A cleaner way to see this series is to split it into two separate sub-sequences that run side by side, one for the odd terms and one for the even terms.

1. Write the terms as (first number, second number) pairs against the counting number n = 1, 2, 3, 4, 5, 6, 7: (1,1), (8,4), (9,27), (64,16), (25,125), (?,?), (49,343).
2. Check the odd positions n = 1, 3, 5, 7: the pairs are (1,1), (9,27), (25,125), (49,343). Here $9=3^2$, $27=3^3$; $25=5^2$, $125=5^3$; $49=7^2$, $343=7^3$. So for odd n the rule is $(n^2, n^3)$.
3. Check the even positions n = 2, 4: the pairs are (8,4), (64,16). Here $8=2^3$, $4=2^2$; $64=4^3$, $16=4^2$. So for even n the rule is $(n^3, n^2)$, exactly the reverse order.
4. The missing pair is at position n = 6, which is even, so it must follow the even-position rule $(n^3, n^2)$.
5. Compute $6^3 = 216$ and $6^2 = 36$, giving the pair $216 : 36$.
6. This matches option (2) exactly. None of the other options are square-cube pairs of 6, so they can be dropped.

\[\boxed{216 : 36}\]