Question

Show that the sequence, defined by its nth term 3+n by 4,forms an AP. Also, find the common

Answer 1

hope it will help u......

Answer 2

The common difference(d) $=\dfrac{1}{4}$.

Step-by-step explanation:

We have,

The sequence of nth term,

$a_{n} =\dfrac{3+n}{4}$

To find, the common difference(d) = ?

Put n = 1, we get

First term of an AP,

$a_{1} =\dfrac{3+1}{4}=\dfrac{4}{4} =1$

Put n = 2, we get

Second term of an AP,

$a_{2} =\dfrac{3+2}{4}=\dfrac{5}{4}$

Put n = 3, we get

Third term of an AP,

$a_{3} =\dfrac{3+3}{4}=\dfrac{6}{4}=\dfrac{3}{2}$

....

We know that,

Common difference(d) = Second term - First term = Third term - Second term

∴ Second term - First term = $\dfrac{5}{4}-1$

$= \dfrac{5-4}{4}=\dfrac{1}{4}$

Also, Third term - Second term$=\dfrac{3}{2}-\dfrac{5}{4}=\dfrac{6-5}{4}=\dfrac{1}{4}$

The common difference of an AP are same, the given sequence are in AP, proved.

Hence, the common difference(d) $=\dfrac{1}{4}$.