Question

Find the sum of all integers 30 to 80, INCLUSIVE. (Is there a formula to do this?)

Answer 1

((30+80)*50)/2

Answer 2

it will be 30, 31, 32,... 80 forms A.P.
Sn=n/2[2a+(n-1)d]
i.e. 50/2[2*30+(50-1)1]
i.e. 25[60+49]
i.e.25*109
=2725

Answer 3

@myko and @priyadabas answers are my answer chocies, but which one is right?

Answer 4

bouth

Answer 5

Nope yours equal 2750

Answer 6

ohh, so then it's mine, :)

Answer 7

-_- not really it could be wrong

Answer 8

you could find halfway between 30 and 80, which is 55
and add 30+80, 31+79,....54+56, 55
that is 25 times 110 +55

Answer 9

=2805

Answer 10

30 31 32 33 34..... 80
80 79 78 77 76 .... 30
----------------------- +
= twice the summ

Answer 11

OMG The answer chcies are:
A. 2725, what @priyadabas got
B. 2750, what @myko got
C. 2805, what @UnkleRhaukus got
D. 2835

Answer 12

SO CONFUSED :(

Answer 13

http://www.wolframalpha.com/input/?i=sum+from+n%3D+30+to+80+of+n

Answer 14

30 to 80, INCLUSIVE

Answer 15

The formula for calculating sum of an A.P. is
Sn=n/2[2a+(n-1)d]
where n=no. of terms
a=first term
d=diff. b/w two terms.

Answer 16

BUT WHO IS RIGHT??? I dont even know the answer...

Answer 17

myko forgot to add the 55, the middle term

Answer 18

Alright, so it leaves down to @priyadabas and @UnkleRhaukus

Answer 19

SO CONFUSED because the answer chocies are:
A. 2725, what @priyadabas got
B. 2750, what @myko got
C. 2805, what @UnkleRhaukus got
D. 2835

Answer 20

priyadabas has left off the last term, 80

Answer 21

So... @UnkleRhaukus is right?

Answer 22

@UnkleRhaukus - U're wrong. Last term is called tn. I'm right.

Answer 23

AH!!! SO CONFUSED!!!

Answer 24

n it's of no use in the sum only difference n no. of terms are used.

Answer 25

@priyadabas can you see an error in my method?

Answer 26

@UnkleRhaukus -U've used the ncert formula- n(n+1)/2
wich is not wrong but it can't be used like that here. It is for the sum of 'n' integers.

Answer 27

((30+80)*51)/2=2805
i made mistake befor, becouse there are 51 terms to sum, not 50

Answer 28

@Spyush

Answer 29

but
\[(30+80)\times50)/2+55=2805\]
\[(30+80)\times51)/2+55=2860\]

Answer 30

55=110/2

Answer 31

now im confused

Answer 32

@UnkleRhaukus's result 2805 including both left and right ends [ 30, 80 ]

Answer 33

It is 2805. You first find the average \[( 30+80)\div2=55\] Then find the number of numbers so \[80-30+1=51\] Now you multiply these two numbers together \[55 \times 510=2805\]

Answer 34

@myko what?

Answer 35

I'm getting 2805. \[\sum_{n=30}^{80}n=\sum_{n=1}^{80}n-\sum_{n=1}^{29}n\]\[\frac{80(81)}{2}-\frac{29(30)}{2}=40(81)-29(15)=3240-435=2805\]

Answer 36

now there is a method i can understand!

Answer 37

and an answer that is correct

Answer 38

I love the formula
\(1+2+3+...+ n = \frac{n}{2}(1+n)\)
Just saying..