Which term of the AP: 3, 15, 27, 39, ... will be 132 more than its 54th term?
65th term
Step 1: Identify the first term and common difference.
The AP is: 3, 15, 27, 39, ...
First term: \(a = 3\).
Common difference: \(d = 15 - 3 = 12\). (Check: \(27 - 15 = 12\), \(39 - 27 = 12\)).
Step 2: Write the general term.
The nth term of an AP is:
\[a_n = a + (n - 1)d\]
So here,
\[a_n = 3 + (n - 1) \cdot 12 = 3 + 12n - 12 = 12n - 9\]
Step 3: Find the 54th term.
\[a_{54} = 12 \cdot 54 - 9\]
\[a_{54} = 648 - 9 = 639\]
Step 4: Set up the condition “132 more than the 54th term”.
If some term \(a_n\) is 132 more than \(a_{54}\), then:
\[a_n = a_{54} + 132\]
Substitute values:
\[12n - 9 = 639 + 132\]
\[12n - 9 = 771\]
Step 5: Solve for \(n\).
\[12n = 771 + 9 = 780\]
\[n = \dfrac{780}{12} = 65\]
Conclusion: The term which is 132 more than the 54th term is the 65th term of the AP.