In this post I’ll share with you a useful shortcut maths trick for **“finding out the sum of consecutive numbers”**. For example, this trick I am talking about can help you in finding the sum of all the numbers from 23 to 31 or any other set of numbers.

## Shortcut Addition Trick

Add the smallest number to the largest number of the given set of consecutive numbers. Then multiply the result by the number of numbers in the set. Finally divide the result by two.

Solving the above example, let’s find: 23+24+25+26+27+28+29+30+31

Step 1: Add the smallest and the largest number from the above set of numbers:

23 + 31 = 54

Step 2: Multiply the result by the number of numbers in the above set. In the above set there are 9 numbers from 23 to 31.

Therefore, multiply 54 by 9

54 x 9 = 486

Step 3: Finally, divide the above result by 2

486/2 = 243

Hence, 23+24+……..+31 = 243

So now, since you know this simple calculation trick, you don’t have to add up each number individually to get the answer. With a little practice, this trick might become a good tool to save lot of your time. In adding numbers the biggest problem faced is the issue of carryover. Check out how carryover in addition can be avoided to make any addition much simpler.

If you want to suggest some additions or modification in the above method, feel free to post your suggestion as comment below. To learn another shortcut addition trick, read this post

I am glad to include the suggestions posted as comment below, for the benefit of everyone.

**Suggestion by Sagar Shah –**

“If there are odd number of terms then multiply the middle term with number of terms and you get the answer.

We will take the same example. 23+24+25+26+27+28+29+30+31

Here the middle term = 27 and the num of terms is 9.

Therefore the answer is 9*27 = 243

If there are even number of terms then take the mean of the two middle terms

eg 23+24+25+26+27+28+29+30

Here there are 8 terms and the two middle terms are 26 and 27. So mean is 26.5. Multiply it with num of terms i.e 8

Solution is 26.5 * 8 = 212″

One very simple formula is used to deduce this addition shortcut. If you could identify that, post it as comment below.

Sir, these are really worthy.

Thanks for update

IN AIRTHMETIC PROGRESSION THE SUM OF N TERM IS

S= N/2(A+L)

HERE N= NO OF TERMS , A= FIRST TERM, L= LAST TERM

EX 2+3+4+5+6+7

HERE N=6 ,A=2 , L=7

S=6/2(2+7) = 3*9 =27

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Thankyou for your tricks. I hope you can become a good mathematician

Patawari sir, all your posts are very useful

useful one. thanks

The sum of 9 consecutive odd numbers of set A is 53.What is the sum of another set of 6 consecutive even nos. whose smallest number is 13 more than the smallest no. of set A.pLEASE TELL BY SHORTCUT METHOD.

Total no of terms = one more than difference of extreme terms.

USEFUL FOR THEM WHO WOULD LIKE TO AVOID COUNTING in case of a very large no of terms !!!

Indeed!! It makes it more simpler to count the number of terms.

Addition of consecutive numbers should be done in seconds. 🙂

How to find the last two digit for any number ‘n’ with power ‘p’

eg :- 7 raise to 876.

Hi deny,

Power cycle of 7 has a very peculiar characteristic.

7 raise to any number will either end in 01,07,49,43.

Hence you can say that power cycle of 7 is 4.

For eg consider the following

7^1 = 07 —> ends in 07

7^2 = 49 —> ends in 49

7^3 = 343 —> ends in 43

7^4 = 2401 —> ends in 01

7^5 = 14807 —> ends in 07

7^6 = 117649 —> ends in 49

7^7 = 823543 —> ends in 43

7^8 = 5764801 —> ends in 01

so on and so forth…

So for your ques 7 raise to 876.

divide 876 by 4 –> it is exactly divisible..

Hence 7 raise to 876 will end in 01. i.e. 01 would be the last two digits.

I hope this would be helpful.

You can get back to me on sagar_shah89@hotmail.com if you have any doubt. 🙂

Hi danny , i had solved your problem, you can refer to http://www.drzero.in for solution.I had given general method to calculate last two digit of any number ‘n’ to power ‘p’.

final question , from where yu have taken the bookmark and share icons

rply fast

Concept behind above shortcut is as follow

Sum of n term of A.P = n * (first term + last term)/2 )

To read faster calender calculation method ,Concept behind common balance puzzle

follow my above drzero.in website

Concept behind above shortcut is as follow

Sum of n term of A.P = n * (first term + last term)/2 )

To read faster calender calculation method ,Concept behind common balance puzzle

follow my http://www.drzero.in website

Sorry boss, but it wasn’t tricky at all….

@sagar…nice tip that was. thank u

@vineet….IS there a way, that the readers can also contribute to the page

Dear Nitish,

Reader can definitely contribute on QuickerMaths.com by mailing me the content on vineetpatawari@gmail.com. If you acceptable quality, I’ll post it here.

Thanks

Vineet

isme kaun sa bada trick hai bhai…..simple A.P series k sum ko alag dhang se bata rahe ho. Ghanta shortcut!!!

Dear SRK,

Isme koi bada trick nahi hai. This is meant for class ix – x students, who are yet not aware of AP-GP.

In any case, can we all try to avoid unnecessary words in our comments. I hope you understand what I mean.

Thanks

Vineet

hi vineet , i am anuj and i like this post as i am in 10th class and not aware of this

Hi,

Addition of consecutive numbers still has a better shortcut.

If there are odd number of terms then multiply the middle term with number of terms and you get the answer.

We will take the same example. 23+24+25+26+27+28+29+30+31

Here the middle term = 27 and the num of terms is 9.

Therefore the answer is 9*27 = 243

If there are even number of terms then take the mean of the two middle terms

eg 23+24+25+26+27+28+29+30

Here there are 8 terms and the two middle terms are 26 and 27. So mean is 26.5. Multiply it with num of terms i.e 8

Solution is 26.5 * 8 = 212

Thanks a lot for your suggestion. I have included it in the post itself.

good