Question

which of the following can be the nth term of an A.P 3n+1, 2n²+3, n³+n give reason​

Answer 1

Answer:

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3n +1

Step-by-step explanation:

In case of arithmetic sequence, the difference between two consecutive terms is equal.

The general term of A.P. is

a_n = a + (n - 1)d

an = a + (n−1)d

We can see that the nth term of A.P represents an equation of straight line. The linear equation is in the form of y = mx +b, here m is slope and b is the y-intercept.

Therefore, we can conclude that nth term of an A.P is a linear function.

Hence, among the given options 3n +1 is linear and hence it represents nth term of an A.P.

Hope this will help you ....

.Thank you ....❤

Answer 2

Only $3n+1$ is an A.P., because it is a linear polynomial. Also, a constant can be an A.P.

For proof see below.

Proof

For three numbers to form an A.P. they must add equally.

To prove, let $n=k-1,k,k+1$ for $k\geq 2$.

(1) $3n+1$

Where $n=k-1$, the given term is $3k-2$.

Where $n=k$, the given term is $3k+1$.

Where $n=k+1$, the given term is $3k+4$.

∴ $d_{1}=d_{2}=3$

∴For any three numbers $k-1,k,k+1$ such that $k\geq2$, there exists an A.P.

∴For every natural number, $3n-1$ is an A.P.

(2) $2n^2+3$

Where $n=k-1$, the given term is $2k^2-4k+5$.

Where $n=k$, the given term is $2k^2+3$.

Where $n=k+1$, the given term is $2k^2+4k+7$.

∴ $d_{1}=4k-2,d_{2}=4k+4$

∴But, $d_{1}\neq d_{2}$. Such A.P. cannot exist.

(3) $n^3+n$

Where $n=k-1$, the given term is $k^3-3k^2+4k-2$.

Where $n=k$, the given term is $k^3+k$.

Where $n=k+1$, the given term is $k^3+3k^2+4k+2$.

∴ $d_{1}=3k^2-3k+2,d_{2}=3k^2+3k+2$

∴But, $d_{1}\neq d_{2}$. Such A.P. cannot exist.