Prove that no matter what the real numbers a and b are,
Question:

Prove that no matter what the real numbers a and are, the sequence with nth term a + nb is always an A.P. What is the common difference?

Solution:

In the given problem, we are given the sequence with the $n^{\text {th }}$ term $\left(a_{n}\right)$ as $a+n b$ where $a$ and $b$ are real numbers.

We need to show that this sequence is an A.P and then find its common difference $(d)$

Here,

$a_{n}=a+n b$

Now, to show that it is an A.P, we will find its few terms by substituting $n=1,2,3$

So,

Substituting n = 1we get

$a_{1}=a+(1) b$

$a_{1}=a+b$

Substituting n = 2we get

$a_{2}=a+(2) b$

$a_{2}=a+2 b$

Substituting n = 3we get

$a_{3}=a+(3) b$

 

$a_{3}=a+3 b$

Further, for the given to sequence to be an A.P,

Common difference $(d)=a_{2}-a_{1}=a_{3}-a_{2}$

Here,

$a_{2}-a_{1}=a+2 b-a-b$

$=b$

Also,

$a_{3}-a_{2}=a+3 b-a-2 b$

$=b$

Since $a_{2}-a_{1}=a_{3}-a_{2}$

Hence, the given sequence is an A.P and its common difference is $d=b$.

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