Question

Show that $ {a_1},{a_2},...,{a_n},...$ form an AP where $ {a_n} $ is defined as below:
(i)$ {a_n} = 3 + 4n $
(ii)$ {a_n} = 9 - 5n $

Answer 1

Hint: Find the first, second, and third terms of the sequence and then use them to form a sequence. Then use the fact that the common difference is the same for the difference of each consecutive term, and get the desired result.

Complete step by step solution:
We have given in the part (i) that $ {a_n} = 3 + 4n $.
We have to show that $ {a_1},{a_2},...,{a_n},... $form an AP.
$ {a_n} = 3 + 4n $
Find the first term of the sequence by the substitution$n = 1$, then we have
$ {a_1} = 3 + 4\left( 1 \right) $
$ {a_1} = 3 + 4 $
$ {a_1} = 7 $
So, the first term of the sequence is $ {a_1} = 7 $.
Now, find the second and the third terms of the sequence by the substitution of $ n = 2 $ and $ n = 3 $.
$ {a_2} = 3 + 4\left( 2 \right )$
$ {a_2} = 3 + 8 $
$ {a_2} = 11 $
$ {a_3} = 3 + 4\left( 3 \right) $
$ {a_3} = 3 + 12 $
$ {a_3} = 15 $
Thus, the second and the third term of the sequence are $11$ and $15$.
So, the series has the form:
$ 7,11,15 \cdot \cdot \cdot {a_n} $
Any series is in A.P. if the common differences between its consecutive terms are the same. That is,
$ {a_2} - {a_1} = {a_3} - {a_2} = {a_{n + 1}} - {a_n} $
This series is an AP if the common difference is the same for any two consecutive terms.
We have,
$ {a_1} = 7,{a_2} = 11,{a_3} = 15 $ and so on.
Now, find the common difference between the second and the first term.
$ {a_2} - {a_1} = 11 - 7 $
$ {a_2} - {a_1} = 4 $
Find the common difference between the third and the second term.
$ {a_3} - {a_2} = 15 - 11 $
$ {a_3} - {a_2} = 4 $
We have to find that the common difference in both the cases is the same, thus the formed series are in AP.
Therefore, the series
$ 7,11,15 \cdot \cdot \cdot {a_n} $ is an AP series.

We have given in the part (i) that ${a_n} = 9 - 5n $.
We have to show that $ {a_1},{a_2},...,{a_n},...$ form an AP.
$ {a_n} = 9 - 5n $ $ {a_n} = 3 + 4n $
Find the first term of the sequence by the substitution$ n = 1 $, then we have
$ {a_1} = 9 - 5\left( 1 \right) $
$ {a_1} = 9 - 5 $
$ {a_1} = 4 $
So, the first term of the sequence is $ {a_1} = 4 $.
Now, find the second and the third terms of the sequence by the substitution of $ n = 2 $ and $ n = 3 $.
$ {a_2} = 9 - 5\left( 2 \right) $
$ {a_2} = 9 - 10 $
$ {a_2} = - 1 $
$ {a_3} = 9 - 5\left( 3 \right) $
$ {a_3} = 9 - 15 $
$ {a_3} = - 6 $
Thus, the second and the third term of the sequence are $ - 1 $ and $ - 6 $.
So, the series has the form:
$ 4, - 1, - 6 \cdot \cdot \cdot {a_n} $
Any series is in A.P. if the common differences between its consecutive terms are the same. That is,
$ {a_2} - {a_1} = {a_3} - {a_2} = {a_{n + 1}} - {a_n} $
This series is an AP, if the common difference is the same for any two consecutive terms.
We have,
$ {a_1} = 4,{a_2} = - 1,{a_3} = - 6 $ and so on.
Now, find the common difference between the second and the first term.
$ {a_2} - {a_1} = - 1 - 4 $
$ {a_2} - {a_1} = - 5 $
Find the common difference between the third and the second term.
$ {a_3} - {a_2} = - 6 - \left( { - 1} \right) $
$ {a_3} - {a_2} = - 5 $

Note: For the A.P. series the difference between each consecutive term has the same value and for the GP series the common ratio has the same value. We use this fact to show that the series is AP or GP.