Find \(a_{20}\) if \(a_1=12\) and \(a_7\) is 24 less than \(a_{11}\).
\(a_{20}=126\)
Step 1: Recall the general formula for the \(n^{th}\) term of an arithmetic progression (AP):
\(a_n = a_1 + (n-1)\,d\)
Here, \(a_1 = 12\) and \(d\) is the common difference (unknown).
Step 2: Write expressions for \(a_7\) and \(a_{11}\):
\(a_7 = a_1 + (7-1)\,d = 12 + 6d\)
\(a_{11} = a_1 + (11-1)\,d = 12 + 10d\)
Step 3: We are told that \(a_7\) is 24 less than \(a_{11}\). That means:
\(a_{11} - a_7 = 24\)
Step 4: Substitute the values we found:
\((12 + 10d) - (12 + 6d) = 24\)
\(12 + 10d - 12 - 6d = 24\)
\(4d = 24\)
Step 5: Solve for \(d\):
\(d = \dfrac{24}{4} = 6\)
Step 6: Now find \(a_{20}\):
\(a_{20} = a_1 + (20-1)\,d\)
\(= 12 + 19 \times 6\)
\(= 12 + 114\)
\(= 126\)
Final Answer: \(a_{20} = 126\)