If \(S_6=36\) and \(S_{16}=256\), find \(S_{10}\).
\(S_{10}=100\)
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\) \]
Here, \(a\) = first term, \(d\) = common difference, \(n\) = number of terms.
Step 2: Write the equation for \(S_6 = 36\).
\[ S_6 = \dfrac{6}{2}\big(2a + (6-1)d\) = 36 \]
Simplify:
\[ 3(2a + 5d) = 36 \]
\[ 2a + 5d = 12 \quad ...(1) \]
Step 3: Write the equation for \(S_{16} = 256\).
\[ S_{16} = \dfrac{16}{2}\big(2a + (16-1)d\) = 256 \]
Simplify:
\[ 8(2a + 15d) = 256 \]
\[ 2a + 15d = 32 \quad ...(2) \]
Step 4: Subtract (1) from (2) to find \(d\).
\[ (2a + 15d) - (2a + 5d) = 32 - 12 \]
\[ 10d = 20 \]
\[ d = 2 \]
Step 5: Put the value of \(d\) into (1) to find \(a\).
\[ 2a + 5(2) = 12 \]
\[ 2a + 10 = 12 \]
\[ 2a = 2 \quad \Rightarrow \quad a = 1 \]
Step 6: Use the formula again to find \(S_{10}\).
\[ S_{10} = \dfrac{10}{2}\big(2a + (10-1)d\) \]
Substitute \(a=1\), \(d=2\):
\[ S_{10} = 5(2(1) + 9(2)) \]
\[ S_{10} = 5(2 + 18) = 5 \times 20 = 100 \]
Final Answer: \(S_{10} = 100\)