Inductive reasoning (Part - 1) - Mathematics, Engineering Video Lecture | Logical Reasoning for UGC NET

What is the nth number in
the following sequence?
Let's see.
The first term here is a 6.
So over here I'll
put the term.
So this is the number.
I'll just write num there.
That's the number.
And this is the term.
So the first term here is a 6.
It looks like we add 3 to
that to get to a 9.
Or maybe there's some
other pattern here.
So that's the second
term, is 9.
Then they go from 9 to 12.
Looks like we add 3 again.
So we keep adding 3.
So the third term is 12,
the fourth term is 15.
And we need to think of a
general expression that will
give us the value
of the nth term.
The nth number in the
following sequence.
So let's think a little
bit about it.
We know that we're incrementing
by 3 every time
and we're starting at 6.
So let's write these in terms
of multiples of 3, since we
know that that's the
increment level.
This first term right
here is 3 times 2.
The second term right
here is 3 times 3.
This third term is 3 times 4.
And so it looks like it's 3
times one more than the actual
term you are dealing with.
Right?
Here, if you take your term,
you add 1 and then multiply
that by 3, you get 6.
Take your term, add 1, then
multiply it by 3.
Take your term, add
1, you get 4.
Then multiply by 3.
Take your term, add 1-- which
would be 5-- and then multiply
by 3, you get 15.
So it looks like any time-- so
if you have your nth term.
So if you go all the way out
here to your nth term, the
value here would be
you'd add 1 to it.
So you'd have n plus 1, and then
you'd multiply that by 3.
So the nth term is
3 times n plus 1.
Or if you wanted to distribute
out the 3, it'd be 3n plus 3.
And just make sure that
works for all of them.
The first term, 3 times 1
is 3 plus 3, that works.
That's 6.
The second term, 3 times
2 plus 3 is 9.
So it works for all
of these terms. So
that is the nth term.
We have to assume that this
pattern continues.