Show that \(a_1, a_2, \ldots, a_n, \ldots\) form an AP where \(a_n\) is defined as below :
(i) \(a_n = 3 + 4n\)
(ii) \(a_n = 9 - 5n\)
Also find the sum of the first 15 terms in each case.
(i) \(S_{15} = 525\)
(ii) \(S_{15} = -465\)
Recall: A sequence is an arithmetic progression (AP) if the difference between any term and its previous term is constant: \(a_{n+1} - a_n = d\). The sum of the first \(n\) terms is
\[S_n = \frac{n}{2}\bigl(2a_1 + (n - 1)d\bigr) = \frac{n}{2}(a_1 + a_n).\]
Step 1: Show it is an AP.
First few terms:
\[a_1 = 3 + 4(1) = 7, \quad a_2 = 3 + 4(2) = 11, \quad a_3 = 3 + 4(3) = 15, \ldots\]
Check the common difference:
\[a_2 - a_1 = 11 - 7 = 4, \quad a_3 - a_2 = 15 - 11 = 4\]
In general,
\[a_{n+1} - a_n = [3 + 4(n + 1)] - [3 + 4n] = 4.\]
Since the difference is always 4, the sequence forms an AP with
\[a_1 = 7, \quad d = 4.\]
Step 2: Find \(S_{15}\).
Use \(S_n = \frac{n}{2}[2a_1 + (n - 1)d]\) with \(n = 15\):
\[S_{15} = \frac{15}{2}[2\cdot 7 + (15 - 1)\cdot 4]\]
\[= \frac{15}{2}[14 + 14\cdot 4]\]
\[= \frac{15}{2}[14 + 56] = \frac{15}{2} \cdot 70\]
\[= 15 \cdot 35 = 525.\]
So, \(S_{15} = 525\).
Step 1: Show it is an AP.
First few terms:
\[a_1 = 9 - 5(1) = 4, \quad a_2 = 9 - 5(2) = -1, \quad a_3 = 9 - 5(3) = -6, \ldots\]
Check the common difference:
\[a_2 - a_1 = -1 - 4 = -5, \quad a_3 - a_2 = -6 - (-1) = -5\]
In general,
\[a_{n+1} - a_n = [9 - 5(n + 1)] - [9 - 5n] = -5.\]
So the sequence is an AP with
\[a_1 = 4, \quad d = -5.\]
Step 2: Find \(S_{15}\).
First find the 15th term:
\[a_{15} = 9 - 5(15) = 9 - 75 = -66.\]
Now use \(S_n = \frac{n}{2}(a_1 + a_n)\):
\[S_{15} = \frac{15}{2}(4 + (-66)) = \frac{15}{2}(-62)\]
\[= 15 \cdot (-31) = -465.\]
So, \(S_{15} = -465\).