If $a_{n} = 3 + 4 n$ , show that $a_{1}, a_{2}, a_{3}, a_{4} \dots \dots . . a_{n}$ , from an arithmetic progression. Also find the sum of first $25$ terms.

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Solution

Step 1: Find the terms of the given sequence
Given, $a_{n} = 3 + 4 n$
Substitute $n = 1$ in given equation
$a_{1} = 3 + 4 (1) a = 7$
Substitute $n = 2$ in given equation
$a_{2} = 3 + 4 (2) a_{2} = 11$
Substitute $n = 3$ in given equation
$a_{3} = 3 + 4 (3) a_{3} = 15$
Step 2: Identify the sequence as an arithmetic progression
As we can see $7, 11, 15 \dots$ form the given sequence.
A sequence is considered to be an arithmetic progression when the difference between two successive terms is constant.
Here the difference between two consecutive terms is $15 - 11 = 11 - 7 = 4$
So, the given sequence $a_{n} = 3 + 4 n$ is an arithmetic progression
Step 3: Find the sum of first $25$ terms
The sum of first $n$ terms of an arithmetic progression is given as
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
We have $n = 25, a = 7, d = 4$
Substituting these values in the standard result we get
$S_{n} = \frac{25}{2} [2 \times 7 + (25 - 1) 4]$
$= 25 (7 + 24 \times 2) = 25 (55)$
$\Rightarrow S_{n} = 1375$
Hence the sum of the first $25$ terms is $1375$ .

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