Question

If the sum of the first $ n $ terms of an arithmetic progression is given by $ S_n = 3n^2 + 5n $, find the first term $ a $ and common difference $ d $.

• A. \( a=8, d=6 \)
• B. \( a=8, d=3 \)
• C. \( a=5, d=6 \)
• D. \( a=3, d=5 \)

Hint:

Tip: Remember, \( t_n = S_n - S_{n-1} \) is useful to find terms from sum expressions.

Answer 1

The Correct Option is A

Given the sum of the first n terms of an arithmetic progression (AP): \( S_n = 3n^2 + 5n \), we need to determine the first term \( a \) and the common difference \( d \). 

Step 1: Use Formulas for Sum of AP

The sum \( S_n \) of the first \( n \) terms of an AP is given by:

\( S_n = \frac{n}{2}(2a + (n-1)d) \)

Equating the given expression for \( S_n \):

\( \frac{n}{2}(2a + (n-1)d) = 3n^2 + 5n \)

Multiplying both sides by 2 to clear the fraction gives:

\( n(2a + (n-1)d) = 6n^2 + 10n \)

Simplifying, divide by \( n \):

\( 2a + (n-1)d = 6n + 10 \)

Step 2: Find \( a \) and \( d \)

Setting \( n = 1 \) to find \( a \):

\( 2a + (1-1)d = 6(1) + 10 \)

\( 2a = 16 \)

\( a = 8 \)

For \( n = 2 \):

\( 2a + (2-1)d = 6(2) + 10 \)

\( 2(8) + d = 22 \)

\( 16 + d = 22 \)

\( d = 6 \)

Solution: The first term \( a \) is 8, and the common difference \( d \) is 6.

This matches the option \((a=8, d=6)\).

Answer 2

Step 1: Use formula for \( S_n \) 
Given, \( S_n = 3n^2 + 5n \).

Step 2: Find first term \( a \) 
First term \( a = S_1 = 3(1)^2 + 5(1) = 3 + 5 = 8 \).

Step 3: Find second term \( t_2 \) 
\[ t_2 = S_2 - S_1 = \big(3(2)^2 + 5(2)\big) - 8 = (12 + 10) - 8 = 14 \]

Step 4: Calculate common difference \( d \) 
\[ d = t_2 - a = 14 - 8 = 6 \]