The nth term of a sequence is na+b. Prove that the sequence is an A. P. with common difference .
Answers
Here we should imagine n taking on the values 1,2,3,4,….
So the first term (when n=1) is na+b=1a+b=a+b.
The second term (when n=2) is na+b=2a+b.
The third term (when n=3) is na+b=3a+b, etc.
So the sequence will be [a+b,2a+b,3a+b,4a+b,…]
Well, we can write out the first few terms of the sequence, when n=1,2,3,4 say. (see at the top for how we get this). That tells us the sequence is
[a+b,2a+b,3a+b,4a+b,…]
What next? One of the things we have to do is prove the common difference is a. Let's try looking at some of the differences.
The difference between the first and second terms is (2a+b)−(a+b)=2a+b−a−b=a. Hey, that's good!
The difference between the second and third terms is (3a+b)−(2a+b)=3a+b−2a−b=a.
The difference between the third and fourth terms is (4a+b)−(3a+b)=4a+b−3a−b=a.
At this point you can see this pattern will continue.
You could write your proof as:
The sequence is [a+b,2a+b,3a+b,4a+b,…]
The difference between the first and second terms is (2a+b)-(a+b) = a
The difference between the second and third terms is (3a+b)-(2a+b) = a
The difference between the third and fourth terms is (4a+b)-(3a+b) = a
Therefore the sequence is an AP with common difference a.
Answer: the first term (when n=1) is na+b=1a+b=a+b.
The second term (when n=2) is na+b=2a+b.
The third term (when n=3) is na+b=3a+b, etc.
the sequence will be [a+b,2a+b,3a+b,4a+b,…]
Well, we can write out the first few terms of the sequence, when n=1,2,3,4 say. (see at the top for how we get this). That tells us the sequence is
[a+b,2a+b,3a+b,4a+b,…]
What next? One of the things we have to do is prove the common difference is a. Let's try looking at some of the differences.
The difference between the first and second terms is (2a+b)−(a+b)=2a+b−a−b=a. Hey, that's good!
The difference between the second and third terms is (3a+b)−(2a+b)=3a+b−2a−b=a.
The difference between the third and fourth terms is (4a+b)−(3a+b)=4a+b−3a−b=a.
You could write your proof as:
The sequence is [a+b,2a+b,3a+b,4a+b,…]
The difference between the first and second terms is (2a+b)-(a+b) = a
The difference between the second and third terms is (3a+b)-(2a+b) = a
The difference between the third and fourth terms is (4a+b)-(3a+b) = a
Therefore the sequence is an AP with common difference a.