Find the 12th term from the end of AP: \(-2,-4,-6,\ldots,-100\).
\(-78\)
Step 1: First, note that the sequence is an Arithmetic Progression (AP): \(-2, -4, -6, ...\).
Here, the first term \(a = -2\), and the common difference \(d = -4 - (-2) = -2\).
Step 2: Find the total number of terms in the AP.
We know the last term \(l = -100\).
Formula for the \(n^{th}\) term: \(a_n = a + (n-1) \times d\).
Put the values: \(-100 = -2 + (n-1)(-2)\).
\(-100 = -2 - 2(n-1)\).
\(-100 = -2 - 2n + 2\).
\(-100 = -2n\).
So, \(n = 50\).
This means there are 50 terms in the AP.
Step 3: To find the 12th term from the end, we use:
Position from start = \(n - 12 + 1 = 50 - 12 + 1 = 39\).
So, the 12th term from the end is the 39th term from the beginning.
Step 4: Find the 39th term.
Formula: \(a_{39} = a + (39-1) \times d\).
\(a_{39} = -2 + 38 \times (-2)\).
\(a_{39} = -2 - 76\).
\(a_{39} = -78\).
Final Answer: The 12th term from the end is \(-78\).