# Write the sequence with nth term:

Question:

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.

Solution:

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 = 2we get

$a_{2}=3+4(2)$

$a_{2}=11$

Substituting n = 3we 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 = 2we get

$a_{2}=5+2(2)$

$a_{2}=9$

Substituting n = 3we 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 = 2we get

$a_{2}=6-2$

$a_{2}=4$

Substituting n = 3we 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 = 2we get

$a_{2}=9-5(2)$

$a_{2}=-1$

Substituting n = 3we 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.