Write the sequence with nth term:
(i) $a_{n}=3+4 n$
(ii) $a_{n}=5+2 n$
(iii) $a_{n}=6-n$
(iv) $a_{n}=9-5 n$
Show that all of the above sequences form A.P.
In the given problem, we are given the sequence with the $n^{\text {th }}$ term $\left(a_{n}\right)$.
We need to show that these sequences form an A.P
(i) $a_{n}=3+4 n$
Now, to show that it is an A.P, we will first find its few terms by substituting $n=1,2,3$
So,
Substituting $n=1$, we get
$a_{1}=3+4(1)$
$a_{1}=7$
Substituting n = 2, we get
$a_{2}=3+4(2)$
$a_{2}=11$
Substituting n = 3, we get
$a_{3}=3+4(3)$
$a_{3}=15$
Further, for the given sequence to be an A.P,
Common difference $(d)=a_{2}-a_{1}=a_{3}-a_{2}$
Here,
$a_{2}-a_{1}=11-7$
$=4$
Also,
$a_{3}-a_{2}=15-11$
$=4$
Since $a_{2}-a_{1}=a_{3}-a_{2}$
Hence, the given sequence is an A.P
(ii) $a_{n}=5+2 n$
Now, to show that it is an A.P, we will find its few terms by substituting $n=1,2,3$
So,
Substituting $n=1$, we get
$a_{1}=5+2(1)$
$a_{1}=7$
Substituting n = 2, we get
$a_{2}=5+2(2)$
$a_{2}=9$
Substituting n = 3, we get
$a_{3}=5+2(3)$
$a_{3}=11$
Further, for the given to sequence to be an A.P,
Common difference $(d)=a_{2}-a_{1}=a_{3}-a_{2}$
Here,
$a_{2}-a_{1}=9-7$
$=2$
Also,
$a_{3}-a_{2}=11-9$
$=2$
Since $a_{2}-a_{1}=a_{3}-a_{2}$
Hence, the given sequence is an A.P
(iii) $a_{n}=6-n$
Now, to show that it is an A.P, we will find its few terms by substituting $n=1,2,3$
So,
Substituting $n=1$, we get
$a_{1}=6-1$
$a_{1}=5$
Substituting n = 2, we get
$a_{2}=6-2$
$a_{2}=4$
Substituting n = 3, we get
$a_{3}=6-3$
$a_{3}=3$
Further, for the given to sequence to be an A.P,
Common difference $(d)=a_{2}-a_{1}=a_{3}-a_{2}$
Here,
$a_{2}-a_{1}=4-5$
$=-1$
Also,
$a_{3}-a_{2}=3-4$
$=-1$
Since $a_{2}-a_{1}=a_{3}-a_{2}$
Hence, the given sequence is an A.P
(iv) $a_{n}=9-5 n$
Now, to show that it is an A.P, we will find its few terms by substituting $n=1,2,3$
So,
Substituting $n=1$, we get
$a_{1}=9-5(1)$
$a_{1}=4$
Substituting n = 2, we get
$a_{2}=9-5(2)$
$a_{2}=-1$
Substituting n = 3, we get
$a_{3}=9-5(3)$
$a_{3}=-6$
Further, for the given sequence to be an A.P,
Common difference $(d)=a_{2}-a_{1}=a_{3}-a_{2}$
Here,
$a_{2}-a_{1}=-1-4$
$=-5$
Also,
$a_{3}-a_{2}=-6-(-1)$
$=-5$
Since $a_{2}-a_{1}=a_{3}-a_{2}$
Hence, the given sequence is an A.P.
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