The sum of first 16 terms of the AP: \(10,6,2,\ldots\) is
−320
320
−352
−400
Step 1: Identify the first term (\(a\)) and the common difference (\(d\)).
The first term is \(a = 10\).
The numbers are decreasing by 4 each time (\(6 - 10 = -4\)).
So, \(d = -4\).
Step 2: Recall the formula for the sum of first \(n\) terms of an AP:
\[ S_n = \dfrac{n}{2} [2a + (n-1)d] \]
Step 3: Put the values into the formula:
Here, \(n = 16\), \(a = 10\), and \(d = -4\).
\[ S_{16} = \dfrac{16}{2} [2(10) + (16-1)(-4)] \]
Step 4: Simplify step by step:
\( \dfrac{16}{2} = 8 \)
Inside the brackets: \( 2(10) = 20 \)
\( (16-1)(-4) = 15 \times -4 = -60 \)
So, inside the brackets: \(20 - 60 = -40\)
Step 5: Multiply:
\( S_{16} = 8 \times (-40) = -320 \)
Final Answer: \(-320\), which is Option A.