First term is \(-5\), last term is \(45\), and sum of the AP is \(120\). Find the number of terms and the common difference.
\(n=6,\; d=10\)
Step 1: Write down the given values.
Step 2: Use the formula for the sum of an AP.
The sum of \(n\) terms of an AP is:
\(S = \dfrac{n}{2}(a + l)\)
Step 3: Substitute the known values.
\(120 = \dfrac{n}{2}(-5 + 45)\)
\(120 = \dfrac{n}{2} \times 40\)
\(120 = 20n\)
Step 4: Solve for \(n\).
\(20n = 120\)
\(n = \dfrac{120}{20} = 6\)
So, the number of terms is \(n = 6\).
Step 5: Use the formula for common difference.
The formula connecting the last term is:
\(l = a + (n - 1)d\)
Step 6: Substitute values.
\(45 = -5 + (6 - 1)d\)
\(45 = -5 + 5d\)
Step 7: Solve for \(d\).
\(45 + 5 = 5d\)
\(50 = 5d\)
\(d = \dfrac{50}{5} = 10\)
Final Answer: Number of terms \(n = 6\), Common difference \(d = 10\).