HI Jaya,

We got (n-1)d=999

Now $$999=3^3\times\ 37$$

Since there are 8 factors, so there are 8 possible solutions.

But when d=999 (n-1)999=999 or n-1=1 or n=2. But there should be atleast three elements in the A.P. but here there are only two terms. Hence we have to remove this case. Thus total number of elements is 8-1=7. Hope this helps. Thanks

Oh okk.. Thank you sir

Hi Naveen,

Since it’s an AP with first term $$1$$ and last term $$1000$$, the relationship between common difference $$d$$ and total terms $$n$$ can be written as following;

$$1000 = 1 + (n-1)d \Rightarrow (n-1)d = 999$$

Thus, since $$n$$ is any integer greater than $$2$$ (at least three terms), and $$d$$ is the common difference which is also an integer, $$(n-1)d$$ is actually nothing but integer*integer.

$$999$$ can be written as $$1*999$$, $$3*333$$, $$9*111$$, …, i.e since $$999= 3^3 \times 37$$, the number of pairs are $$\frac{(3+1)(1+1)} = 8$$

Among these pairs, since $$n\ge 3$$, $$n-1\ge 2$$ and hence $$(1,999)$$ will be rejected. The final answer thus, will be $$7$$.

Hope this helps, feel free to ask if you have any doubts