Question

If the \[{{n}^{th}}\] term of a sequence is an expression of first degree in n, show that it is an AP.

Answer 1

Hint: Assume that the \[{{n}^{th}}\] term of a sequence is an expression of first degree in n is \[pn+q\] . Now, using this get the first term, second term, third term, and fourth term. We know that the common difference of any sequence is the difference between its consecutive terms. Now, get the common difference between the second term and the first term. Similarly, get the common difference between the fourth term and the third term. We know the property that the common difference of any two consecutive terms of an AP is always equal. Now, conclude the answer.

Complete step by step answer:
According to the question, it is given that the \[{{n}^{th}}\] term of a sequence is an expression of first degree in n.
Now, first of all, let us assume that the \[{{n}^{th}}\] term of a sequence is an expression of first degree in n is \[pn+q\] .
The \[{{n}^{th}}\] term of a sequence = \[pn+q\] ……………………………………(1)
Now, let us find the first term of this sequence.
On putting \[n=1\] in equation (1), we get
The first term of the AP = \[p+q\] ……………………………………………..(2)
Similarly, let us find the second term of this sequence.
On putting \[n=2\] in equation (1), we get
The second term of the AP = \[2p+q\] ……………………………………………..(3)
We know that the common difference of any sequence is the difference between its consecutive terms.
Now, on subtracting equation (2) from equation (3), we get
Common difference = Second term-First term = \[\left( 2p+q \right)-\left( p+q \right)=2p-p=p\] ………………………………….(4)
Similarly, let us find the third term of this sequence.
On putting \[n=3\] in equation (1), we get
The third term of the AP = \[3p+q\] ……………………………………………..(5)
Similarly, let us find the fourth term of this sequence.
On putting \[n=4\] in equation (1), we get
The fourth term of the AP = \[4p+q\] ……………………………………………..(6)
Now, on subtracting equation (5) from equation (6), we get
Common difference = Fourth term-Third term = \[\left( 4p+q \right)-\left( 3p+q \right)=4p-3p=p\] ………………………………….(7)
We know the property that the common difference of any two consecutive terms of an AP is always equal …………………………………………(8)
From equation (4) and equation (7), we have the common difference of the sequence and we can see that the common difference is equal to \[p\] in both cases.
Using the property, shown in equation (8), we can say that the assumed sequence whose \[{{n}^{th}}\] is an expression of first degree in n is an Arithmetic progression sequence.
Therefore, the \[{{n}^{th}}\] term of a sequence is an expression of first degree n in AP.

Note:
We can also solve this question by using the formula of the \[{{n}^{th}}\] term of an AP. We know the formula of \[{{n}^{th}}\] term of an AP, \[{{n}^{th}}term=first\,term+common\,difference\left( n-1 \right)\] . Here, in this formula, we can see that \[{{n}^{th}}\] term is an expression of first degree in n, and in the question, it is given that, we have a sequence whose \[{{n}^{th}}\] term is an expression of the first degree. Therefore, the given sequence is in AP.