Question

Find the sum of all natural numbers lying between 100 and 1000 which are the multiples of 5

Answer 1

A number is divisible by 5 if the its last digit is a 0 or 5.

There should be a neat way to do this without counting every single one of them. Let's think.

Answer 2

@Directrix is it 98450 ?

Answer 3

How did you calculate the 98450?

Answer 4

@Directrix t1=100 tn= 195 common difference(d) = 5 no of terms =[(tn-t1)/d]+1

Answer 5

Wow! I just reread the problem and saw that we are no longer counting but finding the sum. I need to start over.

Answer 6

I'm not following your nth term of the series.

Answer 7

i got 99550 is it right ??

Answer 8

@aftabali --> Post your work on this thread, please.

Answer 9

@Directrix S=\[(n/2)(2t1+(n-1)d)\]

Answer 10

i think i might be wrong but work on it as a1=100 and l=1000 n=? d=5 putting these values into the formula l=a1+(n-1)d i got n=181 and by substituting this all in Sn=n/2{2a1+(n-1)d i got this 99550

Answer 11

I don't follow. Would you show the formula with the numbers you used?

Answer 12

l=a1+(n-1)d => 1000=100+(n-1)(d) => 1000=100+5n-5 => 5n=905 and finally n=181 and by using formula Sn=n/2{2a+(n-1)d} we get Sn=181/5{2(100)+(181-1)(5)} => 181/2{2(100)+180(5)}=> 181{100+450}=> 99550 might this work

Answer 13

@JOYAL --> What do you think?

To determine n.

A_n = a_1 + (n-1) d
1000 = 100 + (n-1)5
900 = 5(n-1)
180 = n-1
181 = n

Sum = (n)/2 (a_1 + a_n)
Sum = [(181)/2]* ( 100 + 1000)
Sum = 99 550

Answer 14

@ directrix i did the same one well nice explanation