Question

If we are given a nth term as \[{a_n} = 3 + 4n\] , show that \[{a_1},{a_2},{a_3},{a_4}.......{a_n}\] , form an arithmetic progression. Also find the sum of first \[25\] terms.

Answer 1

Hint: Here, we use some concepts of Arithmetic Progression to solve this problem. We also use the sum of \[n\] terms of an AP formula which is \[{S_n} = \dfrac{n}{2}\left( {2a + (n - 1)d} \right)\] . We will also learn about a progression and all the terminology present in that. And we use another formula, which is \[{a_n} = a + (n - 1)d\] , which gives us the \[{n^{th}}\] term of AP.

Complete step-by-step solution:
 An Arithmetic Progression (AP) is a set of numbers in which the difference between consequent terms is a constant term.
For example, consider a progression \[3,7,11,15,19,......\]
So, here, the difference between a term and its next term is 4 which is a constant. So, this is an AP.
And the constant is known as “common difference” generally represented by \[d\] .
And the first term is denoted by \[a\] .
And \[{n^{th}}\] term is equal to \[a + (n - 1)d\] .
So, finally, generalised AP is written as follows
\[a,(a + d),(a + 2d),.......(a + (n - 1)d).....\]
In the question, the \[{n^{th}}\] term is given as \[{a_n} = 3 + 4n\]
Let us find the \[{(n + 1)^{th}}\] term. To find it, we need to substitute \[n + 1\] in place of \[n\] .
\[ \Rightarrow {a_{n + 1}} = 3 + 4(n + 1)\]
Now let us find the difference between the \[{n^{th}}\] and the \[{(n + 1)^{th}}\] term.
\[ \Rightarrow {a_{n + 1}} - {a_n} = (3 + 4n) - (3 + 4(n - 1))\]
\[ \Rightarrow {a_{n + 1}} - {a_n} = 4n - 4(n - 1)\]
\[ \Rightarrow {a_{n + 1}} - {a_n} = 4\]
So, here the difference between consequent terms is equal to 4, which is a constant.
So, the given sequence \[{a_1},{a_2},{a_3},{a_4}.......{a_n}\] is in AP.
And the sum of first 25 terms is equal to
\[{S_{25}} = \dfrac{{25}}{2}\left( {2a + (25 - 1)d} \right)\]
And here, \[a = {a_1} = 3 + 4(1) = 7\] and \[d = 4\]
\[ \Rightarrow {S_{25}} = \dfrac{{25}}{2}\left( {2(7) + (24)4} \right)\]
\[ \Rightarrow {S_{25}} = \dfrac{{25}}{2}\left( {14 + 96} \right)\]
On simplifying, we get,
\[{S_{25}} = \dfrac{{25}}{2}\left( {110} \right)\]
\[ \Rightarrow {S_{25}} = 1375\]
So, the sum of 25 terms is equal to 1375.

Note: All the numbers in an AP are called as terms. In an AP, the value of the first term can also be a negative value. And similarly, the common difference can also be a negative value. A term of an AP is equal to the average of the terms succeeding and preceding it. For example, if \[{a_1},{a_2},{a_3}\] are in AP, then \[{a_2} = \dfrac{{{a_1} + {a_3}}}{2}\] . We use this formula to solve many problems.