Arithmetic Progression: Finding the Term Number (m)

The problem asks for the value of m given the sum of the first n terms of an Arithmetic Progression (AP) and the value of the m^th term.

Deriving the nth Term Formula

We are given the sum of the first n terms, \(S_n = 3n^2 + 5n\). The formula for the n^th term (\(a_n\)) of an AP can be found using the relationship \(a_n = S_n - S_{n-1}\).

1. Calculate \(S_{n-1}\): Replace n with (n-1) in the sum formula: \( S_{n-1} = 3(n-1)^2 + 5(n-1) \) \( S_{n-1} = 3(n^2 - 2n + 1) + 5n - 5 \) \( S_{n-1} = 3n^2 - 6n + 3 + 5n - 5 \) \( S_{n-1} = 3n^2 - n - 2 \)
2. Calculate \(a_n\): Subtract \(S_{n-1}\) from \(S_n\): \( a_n = S_n - S_{n-1} \) \( a_n = (3n^2 + 5n) - (3n^2 - n - 2) \) \( a_n = 3n^2 + 5n - 3n^2 + n + 2 \) \( a_n = 6n + 2 \) This is the formula for the n^th term of the AP.

Calculating the Value of m

We are given that the m^th term, \(a_m\), is 68.

1. Use the \(a_n\) formula for \(a_m\): Substitute m for n in the formula \(a_n = 6n + 2\): \( a_m = 6m + 2 \)
2. Set \(a_m\) equal to the given value: \( 6m + 2 = 68 \)
3. Solve for m: \( 6m = 68 - 2 \) \( 6m = 66 \) \( m = \frac{66}{6} \) \( m = 11 \)

Therefore, the value of m is 11.