\(a_5+a_7=52\) and \(a_{10}=46\). Find the AP.
\(a=1,\; d=5\) (AP: 1, 6, 11, …)
Step 1: Recall the formula for the nth term of an AP:
\(a_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference.
Step 2: Write expressions for the required terms.
Step 3: Use the condition \(a_5 + a_7 = 52\).
\((a + 4d) + (a + 6d) = 52\)
\(2a + 10d = 52\)
Divide both sides by 2: \(a + 5d = 26\) … (Equation 1)
Step 4: Use the condition \(a_{10} = 46\).
\(a_{10} = a + (10-1)d = a + 9d = 46\) … (Equation 2)
Step 5: Solve the two equations.
From Equation (1): \(a + 5d = 26\)
From Equation (2): \(a + 9d = 46\)
Subtract (1) from (2):
\((a + 9d) - (a + 5d) = 46 - 26\)
\(4d = 20 \Rightarrow d = 5\)
Step 6: Find \(a\).
Put \(d = 5\) in Equation (1): \(a + 5(5) = 26\)
\(a + 25 = 26 \Rightarrow a = 1\)
Step 7: Write the AP.
The first term is \(1\) and common difference is \(5\).
So, the AP is: 1, 6, 11, 16, …