find the sum of arthimetic progression....?

Question

A.P. 3+11+19+_ _ _ _ +803 .find the sum.

Answer 1

Use this formula:

Sn=n/2[t1+tn]

where
Sn --> ur required sum
n --> no. of terms in the given progression
t1 --> first term (in ur case: 3)
tn -->last term (in ur case: 803)

Answer 2

You have to do a little deconstruction to get this into something easy to solve, then reconstruct it.

First, subtract 3 from all you numbers to give you 0+8+16+... +800.

Then divide each of your numbers by 8 to give you the sum of 0 to 100.

This is an easy problem to solve. If you pair the first with the last, the second with second to last, etc., you have 500 pairs whose sum is 100, plus an extra 50.

In other words, the sum of 0 to 100 equals n/2*(n+1) where n is your highest number, 100).

Now you need to do a little reconstruction. Each of those numbers should have been 8 times higher, so multiply your sum by 8.

You need to add 3 back in to all of your numbers. You had 101 numbers from 0 to 100 (you have to include the zero), so add in you 3 *101.

That should get you a final answer of 40703.

(If you know how to use excel, including the summing function, you can always do a quick double check, just to make sure you didn't make a mistake along the way.)

Edit: This is essentially the same solution already posted by others, but just explains why they're doing what they're doing.

Answer 3

Use Euler's method:
S = 3+11+19+...803
S = 803 + 795 + 787+...3
2S = 806 * 100= 80600 (we multiply by 100 since there are 100 of the 806's).
So S = 80600/2= 40300

Answer 4

We can see that the first term is 3, last term is 803, and the difference between the succesive terms is 8;
803=3+(n-1)d (where d=difference bet succesive terms,3 is the first term,803 is the last term, nis the no. of terms)
which gives 800/8=n-1
hence n=101
sum of given AP is given by
(a+l)n/2
where a=first term,l is the last term
sum =(3+803)101/2
=40703

Answer 5

3+11+19+_ _ _ _ +803
each term is a(i)= 3 + 8*i for i=1,...,100
for (i=1 to 100)
a(0)=3
a(1)=11
sum(a(i)) =a(0)+ sum(3) + sum(8*i)
sum(a(i))=3+ sum(3) + 8* sum(i)
for i=1 to 100
note:
sum of i=1 to 100 is (101)100/2.
sum(i)=50*101
sum(a(i))=3+ 3*100+8*(50*101)=
303+101*400=101(3+100*4)=
101*(403)=40300+403=
40,703

=>
3+11+19+_ _ _ _ +803=40,703
--------------------------
NOTE-
sum of intergers
i=1,2,3,....,n
is (n+1)*n/2.
--------------------------
example:
sum of 1,2,3,4,5, 6
line up these six terms forward & backward and do the sums
1 2 3 4 5 6
6 5 4 3 2 1
___________
7 7 7 7 7 7
Now you have 6 times 7 divided by 2.
the answer is (6*7)/2=21 for the sums 1,2,3,4,5,6.
Same for i=1 to n.
==========================

Answer 6

Well, each term is 8 higher than its predecessor, so we can write the series as

SUM(8i+3), i from 0 to 100

(Sorry, Yahoo Answers provides no way to type proper "sigma" notation.)

We can simplify like so:

SUM(8i+3) = 8*SUM(i) + SUM(3), i from 0 to 100
= 8*SUM(i) + 3*101, i from 0 to 100
= 8*SUM(i) + 3*101, i from 1 to 100 [at this point, we can ignore the i=0 term]
= 8*100*101/2 + 3*101 [SUM(i) with i from 1 to n = n(n+1)/2]
= 101(8*100/2 + 3)
= 101(403)
= 40703

If there's anything in there you don't follow, please feel free to ask. Hope that helps!

Answer 7

increment=b
number of elements=n
elements are a1, a2,... an
sum=na1+bn(n-1)/2
in your case it's: 101x3+8x101x100/2=40703

Answer 8

40703

Answer 9

40703

Answer 10

40,703