The sum of first \(n\) terms of an AP with \(a=8\), \(d=20\) equals the sum of first \(2n\) terms of another AP with \(a=-30\), \(d=8\). Find \(n\).
\(n=11\)
Step 1: Recall the formula for the sum of first \(n\) terms of an AP:
\[ S_n = \dfrac{n}{2}\Big(2a + (n-1)d\Big) \]
Step 2: Write the sum of the first \(n\) terms of the first AP.
Here, \(a = 8\), \(d = 20\).
\[ S_n = \dfrac{n}{2}\Big(2(8) + (n-1)(20)\Big) \]
\[ S_n = \dfrac{n}{2}\Big(16 + 20n - 20\Big) \]
\[ S_n = \dfrac{n}{2}(20n - 4) \]
\[ S_n = n(10n - 2) \]
Step 3: Write the sum of the first \(2n\) terms of the second AP.
Here, \(a = -30\), \(d = 8\), and the number of terms is \(2n\).
\[ S_{2n} = \dfrac{2n}{2}\Big(2(-30) + (2n-1)(8)\Big) \]
\[ S_{2n} = n\Big(-60 + (2n-1)(8)\Big) \]
\[ S_{2n} = n(-60 + 16n - 8) \]
\[ S_{2n} = n(16n - 68) \]
Step 4: According to the question,
\[ S_n = S_{2n} \]
So,
\[ n(10n - 2) = n(16n - 68) \]
Step 5: Simplify the equation.
Since \(n \neq 0\), divide both sides by \(n\):
\[ 10n - 2 = 16n - 68 \]
Step 6: Solve for \(n\).
\[ -2 + 68 = 16n - 10n \]
\[ 66 = 6n \]
\[ n = 11 \]
Final Answer: \(n = 11\)