Justify whether it is true to say that the following are the $ n^{\text {th }} $ terms of an AP.
$ 2 n-3 $

To do:

We have to justify whether it is true to say that given terms are the \( n^{\text {th }} \) term of an AP.

Solution:

(i) To  check whether the sequence defined by $a_n = 2 n-3$ is an A.P., we have to check whether the difference between any two consecutive terms is equal.

Let us find the first few terms of the sequence by substituting $n=1, 2, 3....$

When $n=1$,

$a_1=2 (1)-3$

$=2-3$

$=-1$

$a_2=2(2)-3$

$=4-3$

$=1$

$a_3=2(3)-1$

$=6-1$

$=5$

$a_4=2(4)-1$

$=8-1$

$=7$

Here,

$a_2-a_1=1-(-1)=1+1=2$

$a_3-a_2=5-1=4$

$a_2-a_1≠a_3-a_2$

Hence, the given sequence is not an A.P. 

(ii) To  check whether the sequence defined by $a_n = 3n^2 + 5$ is an A.P., we have to check whether the difference between any two consecutive terms is equal.

Let us find the first few terms of the sequence by substituting $n=1, 2, 3....$

When $n=1$,

$a_1=3(1)^2+5$

$=3+5$

$=8$

$a_2=3(2)^2+5$

$=3(4)+5$

$=17$

$a_3=3(3)^2+5$

$=3(9)+5$

$=32$

$a_4=3(4)^2+5$

$=3(16)+5$

$=53$

Here,

$a_2-a_1=17-8=9$

$a_3-a_2=32-17=15$

$d=a_4-a_3=53-32=21$

$a_2-a_1≠a_3-a_2≠a_4-a_3$

Hence, the given sequence is not an A.P. 

(iii) To  check whether the sequence defined by $a_n = 1 + n + n^2$ is an A.P., we have to check whether the difference between any two consecutive terms is equal.

Let us find the first few terms of the sequence by substituting $n=1, 2, 3....$

When $n=1$,

$a_1=1+1+(1)^2$

$=1+1+1$

$=3$

$a_2=1+2+(2)^2$

$=3+4$

$=7$

$a_3=1+3+(3)^2$

$=4+9$

$=13$

$a_4=1+4+(4)^2$

$=5+16$

$=21$

Here,

$a_2-a_1=7-3=4$

$a_3-a_2=13-7=6$

$d=a_4-a_3=21-13=8$

$a_2-a_1≠a_3-a_2≠a_4-a_3$

Hence, the given sequence is not an A.P.  

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Updated on: 10-Oct-2022

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