You are not logged in.
Pages: 1
18) the series problem.. here k>=2 if ak = a (k-1) + 4.. and a1= 3.
find the 100th term
11) set A contains multiple of 6 n set b contains multiples of
8. No. of multiplies in set C which has common multiplies of A
and B.
Offline
18.ak-a(k-1)=4 , that means it's a sequence , ak=a1+(k-1)d = 3+(k-1)*4=4k-1=399
dont understand the 11th question ,lol
Last edited by Stanley_Marsh (2006-12-21 16:11:01)
Numbers are the essence of the Universe
Offline
18.ak-a(k-1)=4 , that means it's a sequence , ak=a1+(k-1)d = 3+(k-1)*4=4k-1=399
dont understand the 11th question ,lol
hi stanley
can yu explain how yu got 399
Offline
If I understand your notation currently, this is what you're saying:
That just means that each term is equal to the previous term plus 4. So the second term is 7 (3+4) and the 3rd term is 11 (7+ 4). So to get the 100 term you would have to add 4 ninety times to the first term of 3. 3 + (4*99) = 399.
Offline
excellent, great guys
Offline
Pages: 1