Question

``````       1
3   5
7   9   11
13   15   17   19
21   23 ...........
``````

The extremities of the row containing the number 2015 is ;

a)1981
b)1979
c)2071
d)2069

Here,we will use the concept of arithmetic progression to solve this question.

Approach 1:

To find the extemities of the row. We will first find the row which contains 2015.

Since they are all odd number and they are in AP. And so,by applying the formula for finding the nth term,we have:

p={a+(n-1)d} where p is any term.

That is, 2015={1+(n-1)2}. 2015 is 1008th term in the row whose extremities we have to find.

Now from this we need to find the row which contains the 1008th element. In rth row there are r elements.

So we need r such that 1+2+..+r-1<1008<1+2+..r

From this,we get;
(r-1)r/2 < 1008 < r(r+1)/2

=>(r-1)r < 2016 < r(r+1)

=> r=45

Now you need the first and the last term of the row. Get the equation and solve it.

Approach 2

Now instead of looking at the row and then the extremity, we can say

Left extremity<2015

And left extremity is =1+2+4+..+2r and right extremity is =1+2+4+..+2r+2(r+1)-2

We need to find 1+r(r+1)<2015<(r+1)(r+2)-1

=> r=44

So the extremirty is 1981 and 2069