Question

If the sum of \( n \) terms of an A.P. is \( 3n^2 + 5n \) and its \( m^{th} \) term is \( 164 \), then the value of \( m \) is:

• A. \( 26 \)
• B. \( 27 \)
• C. \( 28 \)
• D. \( 29 \)

Hint:

If \( S_n = An^2 + Bn \), then the common difference is \( d = 2A \) and the first term is \( a_1 = A + B \).

Answer 1

The Correct Option is B

Step 1: Determine the first term (\( a_1 \)).
The sum of the first \( n \) terms is \( S_n = 3n^2 + 5n \).
The first term is found by calculating \( a_1 = S_1 = 3(1)^2 + 5(1) = 3 + 5 = 8 \).

Step 2: Calculate the common difference (\( d \)).
The sum of the first two terms is \( S_2 = 3(2)^2 + 5(2) = 12 + 10 = 22 \). The second term is \( a_2 = S_2 - S_1 = 22 - 8 = 14 \).
The common difference is \( d = a_2 - a_1 = 14 - 8 = 6 \).

Step 3: Derive the general formula for the \( n^{th} \) term (\( a_n \)).
The \( n^{th} \) term is \( a_n = a_1 + (n - 1)d = 8 + (n - 1)6 = 8 + 6n - 6 = 6n + 2 \).

Step 4: Solve for \( m \) using the \( m^{th} \) term.
Given \( a_m = 164 \), we have \( 6m + 2 = 164 \).
\[\n6m = 162\n\]\n\[\nm = \frac{162}{6} = 27\n\]