number pattern in mathematics.....

[hyder]

what is the nth term of the sequence
3,6,10,15,21,......?

 

[abcde]

The nth term is obtained by adding n(n+1)/2 to the previous term..

 

[Chem_master]

at first find the equation of this sequence in terms of x then substitute n from x in the equation....

 

[hyder]

can u give me some tips on this?

 

[Chem_master]

well the equation of this sequence in terms of x is 3x+(x2-x)/x
substitute n with x to achieve answer.....

 

[Nibz]

can u give me some tips on this?

Der r mainly 2 types of sequences viz Linear and quadratic-the latter being more difficult!
Formula for FINDING THE nth TERM OF A LINEAR SEQUENCE = a+(n-1)d here a=the 1st term of the sequence; in our case it is 3 , and 'd' is the difference, here the d = 6) put this in the formula and u will get the nth term!

Now the sequence that you have posted, is called a quadratic sequence i.e the difference b/w 2 consecutive terms, varies!

Your questions asks us to find the nth term of this sequence 1,3,6,10,15......
here the difference b/w 1st 2 consecutive terms is 2 then 3 then 4, 5 and so on and so forth.. *(2,3,4,5,6.....)*
Now we will take difference2 ==> Difference b/w the terms in 2,3,4,5,6... sequence! So Difference2 = 1

Now the formula to find the nth term of a quadratic sequence is " a + (n-1)d1 + 1/2(n-1)(n-2)d2 "
here a = the first term of the sequence, in our case it is 1;
d1= the first term of the first difference i.e 2,
d2 = 1
Substitute these values in the formula!
nth term = 1 + (n-1)2 + 1/2(n-1)(n-2)d2
Solve this in your rough notebook and you will get the answer!!!

 

[hyder]

sory man the first number is not 1 but 3.
i've also chkd the answer frm th ms n it is is (n+1)(n+2)/2
but i dnt undrstand it.....

 

[Nibz]

This is not a BIG DEAL!
if the first term is 3, then we have the value of a = 3, D1 =3 and D2 =1 ; put these values in the formula ==> a + (n-1)d1 + 1/2 (n-1) (n-2)d2
==> 3+(n-1)3 + 1/2 (n-1)(n-2)1
= (3 + 3n - 3) + 1/2 (n^2 - 2n - n + 2)
= (3n) + (n^2 - 3n + 2) /2
= (6n + n^2 - 3n + 2)/2
= (n^2 + 3n + 2)/2
= (n^2 + 2n + n + 2)/2
= (n+1)(n+2)/2

I have done each and every step for you!

 

[hyder]

what is d1 and d2?

 

[Nibz]

what is d1 and d2?

d1 is the first term of the difference u have taken! For your sequence (3,6,10,15....) the first difference is 3,4,5,6....
So D1 is the first term of the first difference i.e 3
D2 is the second difference! 3,4,5,6... differ by only 1 So d2= 1

 

[hyder]

GOT IT!!!!
thnks yr!
*Thumbs up!*

 

[CaptainDanger]

Oh its solved here!!!!

 

[CaptainDanger]

Oh thanks alot NIBz! I did not know the Linear wala formula either... Used the same way as I explained in PM.... This ones easy!!!!!

 

[Nibz]

Not a problem!
And as the cliche' goes, "Time management is more important than anything."

 

[xaadxalman]

thanks nibx .......

 

[CaptainDanger]

Not a problem!
And as the cliche' goes, "Time management is more important than anything."

True.....

 

[shahlakh]

I think n+2n is the formula for this sequence. So by using this formula you can continue the sequence.
Like for example n=1
1+2(1)=3
n=2
2+2(2)=6
and so on...

 

[CaptainDanger]

:?:

Who did you refer that to?

 

[shahlakh]

I am actually referring to the question that the person asked.

 

[CaptainDanger]

Oh ok....