Numbers which are multiple of 2 = 600
Numbers which are multiple of 3 = 400
Numbers which are multiple of 5 = 240

Total = 1240

Common multiples:
Numbers which are multiple of 6 = 200
Numbers which are multiple of 15 = 80
Numbers which are multiple of 10 = 120
Total = 400

Numbers co-prime to 6 or 15 = 1200 - (1240-400) =360
Since either is asked: Add the multiples of 30 = 40

Numbers co-prime to either 6 or 15 = 360+40 = 400

@Puys: Bhai log man ni lag raha tha aapke bina..Isliye aa gaya...Ab pakka bye bye

y shud we add the multiples of 30.. can u explain a bit more on tat.. please

For 6, we've to remove every multiple of 2,3

1200(1-1/2)(1-1/3)=400

For 8, we've to remove every multiple of 3,5
1200(1-1/3)(1-1/5)=640

Coprime to both 6 and 15:

1200(1-1/2)(1-1/3)(1-1/5)=320.

Total: 400+640-320=720.

dude can u tell me y we have to remove every multiple of 3,5 for 8

h T c Says

hey, what is meant by general term of an AP or GP or HP ?

general form is taken when the difference of the nos in the given sequence are in a.p...it suits perfectly for tat particular type of question

can anyone explain clearly how the general term is
(n^2 - n + 2)/2....just coping it from fb...

forget tat.. jus take the simple general formula as an^2+bn+c
now extrapolate to find the zeroth term of the sequence.. it is 1
if n=0; the equation becomes a(0)+b(0)+c=1; therefore c=1
if n=1; the first term of the sequence is actually 1, so a(1)+b(1)+c=1 and as we know tat c=1, a+b has to be zero; a+b=0;a= -b
if n=2; 4a+2b+c=2; as a result 4a+2b=1; b=-1/2; a=1/2
now sub the values in the general formula
this is it.. 1/2(n^2)+(-1/2)n+1
sub the value of required n and get the answer

hope u understand..!!

product of first 50 odd nos...?

wat is the simple way to find the product of first 50 odd nos..?

whenever the common difference is an AP, try to fit in a square.

Let the nth term be an^2+bn+c

extrapolate to find the 0th term of the series. It would be 3.
So, c=3
Put n=1
a+b+3=3; a+b=0
put n=2
4a+2b+3=5
4a+2b=2
2a=2=>a=1, b=-1

The nth term is n^2-n+3

dude.. an^2+bn+c is a general formula right... can the same be used for any sequence of this form..?

wat if the sequence is like this, 3,5,9,15,23.... upto 2000 nos...
the formula will change uh

@bbwi: excellent dude.. but of these two formulae which is the one tat is appropriate..and can u explain more..i cudn understand fully...

guys can u pls tel me how to proceed with these sort of probs in which the intervals of the sequences are in a.p

find the sum of the series.. 1,2,4,7,11,16 ...... for 2000 terms