If \(a_n=3-4n\), show \(a_1,a_2,a_3,\ldots\) form an AP and find \(S_{20}\).
AP with \(d=-4\); \(S_{20}=-780\)
Step 1: Write the general term.
We are given \(a_n = 3 - 4n\).
Step 2: Find the first few terms.
For \(n = 1\): \(a_1 = 3 - 4(1) = 3 - 4 = -1\).
For \(n = 2\): \(a_2 = 3 - 4(2) = 3 - 8 = -5\).
For \(n = 3\): \(a_3 = 3 - 4(3) = 3 - 12 = -9\).
Step 3: Check if it is an AP.
Common difference \(d = a_2 - a_1 = -5 - (-1) = -4\).
Also, \(a_3 - a_2 = -9 - (-5) = -4\).
Since the difference is constant, the sequence is an AP with \(d = -4\).
Step 4: Use the formula for sum of first \(n\) terms of an AP.
Formula: \(S_n = \tfrac{n}{2} \big(2a_1 + (n-1)d\big)\).
Step 5: Substitute the values.
Here, \(a_1 = -1\), \(d = -4\), \(n = 20\).
So, \(S_{20} = \tfrac{20}{2}\big(2(-1) + (20-1)(-4)\big)\).
\(= 10 \big(-2 + 19(-4)\big)\).
\(= 10 \big(-2 - 76\big)\).
\(= 10(-78)\).
\(= -780\).
Final Answer: The sequence is an AP with common difference \(d=-4\), and \(S_{20} = -780\).