formula for adding subsequent numbers?

Answer 1

Say you have to add x-2, x-1, x, x+1 & x+2 so you have to add 5 figures. Centre one is x so multiply x by 5 thats your answer.

If its even i.e. x, x+1, x+2, x+3 then add two centre ones. ie. (x+1 & x+2 in example) now multiply the result by 2 (because we have 4 nos. here so 4/2 = 2 if you have 100 total no. then you have to multiply by 100/2 = 50.)

Answer is ready.
Lets add 1 to 100 nos.

centre ones are 50 & 51, their addition is 101. Now multiply 101 by 50 (as 100/2=50) so it comes to 5050 and thats our answer.

If we have to add 1 to 99 then 50 (centre one) into 99 ie 4950 is the answer.

Answer 2

adding subsequent number is much more easier with Arithmetic progressions.
the formula is
Sn=n/2( 2a(n-1)d)
where, n=is the number of number for whichyou want to find the sum , a= the first term, d= is the difference of the second term- the first term.

for ur knowledge:
The sum of the components of an arithmetic progression is called an arithmetic series.

Calculating the value of an arithmetic series
The value of an arithmetic series consisting of n terms is given by


Intuitively, this formula can be derived by realizing that the sum of the first and last terms in the series is the same as the sum of the second and second to last terms, and so forth, and that there are roughly n / 2 such sums in the series. An often-told story is that Carl Friedrich Gauss discovered this formula when his third grade teacher asked the class to find the sum of the first 100 numbers, and he instantly computed the answer (5050).

A different way to get the result, that avoids the fuzziness of the previous method when the number of terms is odd, is to think in terms of averages. The value of the arithmetic series is the number of terms in the series times the average value of the terms. The average must be (a1 + an) / 2, since the values appear evenly spaced out around around this point on the real number line. Put another way, is constant and equal to (a1 + an) / 2, which corresponds to the fact that successively taking terms from opposite sides of the series gives a constant average, which therefore must be the average of all terms in the series.


Proof of the formula
Express the arithmetic series in two different ways:
Add both sides of the two equations. All terms involving d cancel, and so we're left with:
Rearranging and remembering that an = a1 + (n − 1)d, we get:

Answer 3

Subsequent Numbers

Answer 4

If you want to add nos.1 to 10 or 1to 20 ec, you have a formula for it. (n x n+1)/2.
For example if you want to add numbers 1 to 10.
Then you have to take n=10, so now using the formula
(10 x 10+1)/2 = (10 x 11)/2 = 110/2 = 55 is the answer.

Answer 5

Say you have to add x-2, x-1, x, x+1 & x+2 so you have to add 5 figures. Centre one is x so multiply x by 5 thats your answer.

If its even i.e. x, x+1, x+2, x+3 then add two centre ones. ie. (x+1 & x+2 in example) now multiply the result by 2 (because we have 4 nos. here so 4/2 = 2 if you have 100 total no. then you have to multiply by 100/2 = 50.)

Answer is ready.
Lets add 1 to 100 nos.

centre ones are 50 & 51, their addition is 101. Now multiply 101 by 50 (as 100/2=50) so it comes to 5050 and thats our answer.

If we have to add 1 to 99 then 50 (centre one) into 99 ie 4950 is the answer.

Answer 6

If this is what you mean:
1+2+3+4+5+6+...............+n
then the answer is n(n+1)/2

Answer 7

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Answer 8

haven't you missed something?
is it for A.P. or G.P. or H.P.?

Answer 9

please give full details.