# Justify whether it is true to say

Question:

Justify whether it is true to say that the following are the nth terms of an

AP.

(i) 2n – 3

(ii) 3n2 + 5

(iii) 1 + n + n2

Solution:

(i) Yes, here $a_{n}=2 n-3$

Put $n=1, \quad a_{1}=2(1)-3=-1$

Put $n=2, \quad a_{2}=2(2)-3=1$

Put $n=3, \quad a_{3}=2(3)-3=3$

Put $n=4 . \quad a_{4}=2(4)-3=5$

List of numbers becomes $-1,1,3, \ldots$

Here, $a_{2}-a_{1}=1-(-1)=1+1=2$

$a_{3}-a_{2}=3-1=2$

$a_{4}-a_{3}=5-3=2$

$\because a_{2}-a_{1}=a_{3}-a_{2}=a_{4}-a_{3}=\ldots$

Hence, $2 n-3$ is the $n$th term of an AP.

(ii) No, $\quad$ here $a_{n}=3 n^{2}+5$

Put $n=1, \quad a_{1}=3(1)^{2}+5=8$

Put $n=2, \quad a_{2}=3(2)^{2}+5=3(4)+5=17$

Put $n=3, \quad a_{3}=3(3)^{2}+5=3(9)+5=27+5=32$

So, the list of number becomes $8,17,32, \ldots$

Here, $a_{2}-a_{1}=17-8=9$

$a_{3}-a_{2}=32-17=15$

$\therefore$$a_{2}-a_{1} \neq a_{3}-a_{2}$

Since, the successive difference of the list is not same. So, it does not form an AP.

(iii) No, $\quad$ here $a_{n}=1+n+n^{2}$

Put $n=1, \quad a_{1}=1+1+(1)^{2}=3$

Put $n=2, \quad a_{2}=1+2+(2)^{2}=1+2+4=7$

Put $n=3, \quad a_{3}=1+3+(3)^{2}=1+3+9=13$

So, the list of number becomes $3,7,13, \ldots$

Here, $\quad a_{2}-a_{1}=7-3=4$

$a_{3}-a_{2}=13-7=6$

$\therefore \quad a_{2}-a_{1} \neq a_{3}-a_{2}$

Since, the successive difference of the list is not same. So, it does not form an AP.