Question

Write a rule for the nth term of the sequence:

3, 7, 11, 15...

Answer 1

Answer:

3 + (n - 1)4

Step-by-step explanation:

We have,

3, 7, 11, 15,.....

Now, we must first find what type of a sequence this is.

Let's see if the consecutive terms have a common difference.

7 - 3 = 4

11 - 7 = 4

15 - 11 = 4

Hence, they have a common difference.

So, they are called Arithemetic Progression.

Now,

This sequence starts with 3 and then 4 is being added to get the next digit.

Let's see how,

Let this sequence be a1, a2, a3, a4,........

Now,

a1 = 3 = 3

a2 = 3 + 4 = 7

a3 = 3 + 4 + 4 = 11

a4 = 3 + 4 + 4 + 4 = 15

Here, if you observe closely, in the 1st term 0 time 4 is added, 2nd term 1 time 4 is added, 3rd term 2 times 4 is added, 4th term 3 times 4 is added.

Thus,

nth term will have (n - 1) times 4.

Hence,

a(nth) = 3 + (n - 1)4

In a more generalised way, if the 1st term is 'a' and the common difference is 'd'.

Then,

a(nth) = a + (n - 1)d

This is the general formula for Arithmetic Progression.

But here it is not required.

We must only find the general formula for the sequence

3, 7, 11, 15,.....

Hence, it is 3 + (n - 1)4, where n is each term of the sequence

Hope it helped and believing you understood it........All the best