Need a shorter and faster way of solving

[degreeplus]

If the sum of the consecutive integers from -22 to x, inclusive, is 72, what is the value of x?

(A) 23
(B) 25
(C) 50
(D) 75
(E) 94

the way i solve it is time consuming and i need a way to solve quickly.

Here is how I solved this:

-22-21-20-19-18-17-16-15-14-13-12-11-10-9-8-7-6-5-4-3-2-1+0+1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25=72

or

-253+325=72

By the way, the answer is (B).

 

[stapel]

Yeesh! That's a nasty one.

Okay, we can use some logic to simplify this a bit. You know that the integers -22 through -1 will "cancel out" with the integers 1 through 22, when you add, effectively zeroing out their contribution. So, since -22 + ... + -1 + 0 + 1 + ... + 22 = 0, you're really needing to find the sum 23 + ... + n that equals 72.

From the form of the question, I'm guessing they've given you the formula for the sum of the first "n" whole numbers. I would either use the formula for the sum of the first "n" natural numbers (subtracting the sum up to n = 22 from the sum up to n = x, setting this difference equal to 72, and solving for x), or else just start counting from 23 on up. It's not like 72 is going to be hard to reach, starting with 23.

Just my thoughts....

Eliz.

 

[degreeplus]

Thanks! that reall helped

 

[soroban]

Hello, degreeplus!

Here's a direct method.

If the sum of the consecutive integers from -22 to \(\displaystyle x\), inclusive, is 72,
what is the value of \(\displaystyle x\)?

. . (A) 23 . . (B) 25 . . (C) 50 . . (D) 75 . . (E) 94

We have an arithmetic series: \(\displaystyle \,\underbrace{-22 \,-\,21 \,-\,20\,+\,\cdots\,-1}_{\text{22 terms}}\,+\,\underbrace{0\,+\,1\,+\,2\,+\,\cdots\,+\,x}_{\text{x+1 terms}}\)

. . with first term \(\displaystyle a\,=\,-22\), common difference \(\displaystyle d\,=\,1\), and \(\displaystyle n\:=\:x\,+\,23\) terms.


The sum of an arithmetic series is: \(\displaystyle \:S \;=\;\frac{n}{2}\cdot\left[2a\,+\,d(n\,-\,1)\right]\)

So we have: \(\displaystyle \:\frac{x\,+\,23}{2}\cdot\left[2(-22)\,+\,(1)(x\,+\,22)\right] \:=\:72\)

. . which simplifies to the quadratic: \(\displaystyle \:x^2\,+\,x\,-\,650\:=\:0\)

. . which factors: \(\displaystyle \x\,-\,25)(x\,+\,26)\:=\:0\)

. . and has roots: \(\displaystyle \:x\:=\:25,\,-26\)

Therefore: \(\displaystyle \:x\,=\,\)25

 

[Denis]

Well, since -22 to 22 = 0, then all we need is
3 consecutive numbers from 23: 23+24+25 = 72 :wink:

 

[TchrWill]

If the sum of the consecutive integers from -22 to x, inclusive, is 72, what is the value of x?

the way i solve it is time consuming and i need a way to solve quickly....

Alternatively

As pointed out by the other excellent responses, -22 to +22 cancel out and all you need is the sum of n consecutive integers starting with 23 that sum to 72.

The sum, S, of a set of "n" consecutive numbers starting with a number other than one is given by
S = b(b + 1)/2 - a(a - 1)/2 "a" and "b" being the first and last numbers of the consecutive group.
This may also be written as
S = (b^2 + b - a^2 + a)/2 or
S = (a + b)(b - a + 1)/2

Therefore, S = (23 + b)(b - 22)/2 = 72 from which we get b^2 + b - 650 = 0.

b = [-1+/-sqrt(1 + 2600)]/2 or b = (-1 + 51).2 = 25 making the three nmbers 23, 24 and 25 which sum to 72.