Find the sum of the last 10 terms of the AP: \(8,10,12,\ldots,126\).
1170
Step 1: Identify the first term and common difference.
The AP is \(8, 10, 12, \ldots, 126\).
So, first term \(a = 8\) and common difference \(d = 10 - 8 = 2\).
Step 2: Find the total number of terms in the AP.
The last term is 126. Use the nth term formula:
\(a_n = a + (n-1)d\)
\(126 = 8 + (n-1) \times 2\)
\(126 - 8 = (n-1) \times 2\)
\(118 = 2(n-1)\)
\(n-1 = 59\)
\(n = 60\)
So, there are 60 terms in total.
Step 3: Find which terms are the last 10 terms.
The last 10 terms are from 51st to 60th.
So we need \(a_{51}\) and \(a_{60}\).
Step 4: Find the 51st term.
Formula: \(a_n = a + (n-1)d\)
\(a_{51} = 8 + (51-1) \times 2 = 8 + 100 = 108\).
Step 5: Find the 60th term.
\(a_{60} = 8 + (60-1) \times 2 = 8 + 118 = 126\).
Step 6: Use sum formula for an AP.
Sum of \(n\) terms = \(\dfrac{n}{2}(first + last)\).
Here, \(n = 10\), first = 108, last = 126.
So, Sum = \(\dfrac{10}{2}(108 + 126) = 5 \times 234 = 1170\).
Final Answer: The sum of the last 10 terms is 1170.