Good question

jimmiejaz wrote:A sequence is defined as follows:
An= n/n+1

How many of the first 100 terms of this sequence are less than 0.891?

7
8
9
10
12

Let's jot down the first few terms to understand the sequence.

if n=1, we get 1/2
if n=2, we get 2/3
if n=3, we get 3/4.

So we need to find the first fraction in the sequence that's greater than .891.

Let's start with choice (a).

7/8 = .875... still under
8/9 = .888.. not there yet!
9/10 = .9. Yay, we're past our mark. Since our fractions are getting larger and larger, we now know that only the first 8 terms are less than .891. Choose (B).

Thanks for the explantion.
I got this question because i started with A0.
How will be sure that this sequence doesnt start from 0?

jimmiejaz wrote:Thanks for the explantion.
I got this question because i started with A0.
How will be sure that this sequence doesnt start from 0?

Sets start with the 1st term, not the 0th term (which doesn't make any sense ).

Stuart Kovinsky wrote:

jimmiejaz wrote:Thanks for the explantion.
I got this question because i started with A0.
How will be sure that this sequence doesnt start from 0?

Sets start with the 1st term, not the 0th term (which doesn't make any sense ).

Stuart , am still not able to follow this. When we study A.P or a G.P we start with A0. If, A0=1, An=n/2(2A0+(n-1)d).

Easy way is to start from the value 0.89. This means 9/10 approx right. n = 9, and n+1 = 10. So, the first 8 terms.

Stuart , am still not able to follow this. When we study A.P or a G.P we start with A0. If, A0=1, An=n/2(2A0+(n-1)d).

I could be mistaken here but I have always seen in an AP

Sn = n/2(2*a1 + (n-1)d)

a1->first term

If it was 0

then an=a+(n-1)d will be a0 = a-d which will not make sense IMO.

my bad!!!!!!
U r absolutely true.
A0 cant be A0-d the eqn will not hold.
I got confused big time.....
Thanks for making it clear.

N/N+1 > 891/1000

1000N>891N+891

108N>891

N>8,blablabla

So N(max) = 8

Choose (B)