Match List I with List II: Here is the formatted version with line breaks in List I: \[ \begin{array}{|c|l|} \hline \text{List I} & \text{List II} \\ \hline \text{a. If the 4th term of a G.P. is the square of its 2nd term,} & \text{I. } 5 \\ \text{ and the first term is 3, then the common ratio is} & \\ \hline \text{b. The first term of an A.P. is 5, the last term is 45,} & \text{II. } 3 \\ \text{ and the sum of the terms is 400. The number of terms is} & \\ \hline \text{c. The sum of three numbers in A.P. is 27,} & \text{III. } 16 \\ \text{and the sum of their squares is 293. The common difference is} & \\ \hline \text{d. The 4th and 54th terms of an A.P. are 64 and -61, respectively.} & \text{IV. } -\frac{5}{2} \\ \text{ The common difference is} & \\ \hline \end{array} \] Choose the correct answer from the options given below.
(a) With \(a = 3\), the condition \(T_4 = (T_2)^2\) gives \(3r^3 = (3r)^2 = 9r^2\), so \(r = 3\) → II.
(b) \(S_n = \frac{n}{2}(a + l)\): \(400 = \frac{n}{2}(5 + 45)\), so \(n = 16\) → III.
(c) Let the numbers be \(9-d\), \(9\), \(9+d\) (their sum is 27). Then \((9-d)^2 + 81 + (9+d)^2 = 293\), so \(243 + 2d^2 = 293\), \(d^2 = 25\), and \(d = 5\) → I.
(d) \(a + 3d = 64\) and \(a + 53d = -61\). Subtracting, \(50d = -125\), so \(d = -\frac{5}{2}\) → IV.
So A-II, B-III, C-I, D-IV.
Answer: (C)
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