Question

Given the sequence: 1, 1/2, 2/2, 1/3, 2/3, 3/3, 1/4, 2/4, 3/4, 4/4, ...

Answer 1

$\\ \sf\longmapsto 1,\dfrac{1}{2},\dfrac{2}{2},\dfrac{1}{3},\dfrac{2}{3},\dfrac{3}{3},\dfrac{1}{4},\dfrac{2}{4},\dfrac{3}{4},\dfrac{4}{4}\dots$

First observe the sequence and simplify .

$\\ \sf\longmapsto 1,\left(\dfrac{1}{2},\dfrac{2}{2}\right),\left(\dfrac{1}{3},\dfrac{2}{3},\dfrac{3}{3}\right),\left(\dfrac{1}{4},\dfrac{2}{4},\dfrac{3}{4},\dfrac{4}{4}\right)\dots$

What we observed?

If you observe the denominator and terms the below relation you get:-.

• If denominator be n then at n the term of that pair the value is 1.

Ex:-

For n=2

The 2nd term of the pair is 1

For n=3

The third term of the pair is 1

For n=4

the fourth term of the pair is 1.

And so on

Now

Let's build a sequence

1=1

2=2+1=3

3=3+2+1=3+3=6

4=4+6=10

5=5+10=15

6=6+15=21

7=7+21=28

8=8+28=36

9=9+36=45

10=10+45=55

11=11+55=66

12=12+66=78

________________________

m th term is 5/11

• Observe the sequence build above.

The sequence pair of 11 lies between 55 and 66 .

Now

as per n rule (noted first)

• This pair has 11 terms
• So 5/11 will be the 5th term of this pair

Now

$\\ \sf\longmapsto m=55+5$

$\\ \sf\longmapsto\boxed{\bf{ m=60}}$

Solution:-2:-

Now observe the built sequence .

• 45<48<55

Now

48th term will lie between the pair of 10

Now

• for n=10 the pair has 10terms .

If we subtract the ending of 9th pair from 48 we get

$\\ \sf\longmapsto 48-45$

$\\ \sf\longmapsto 3$

We came to know that 48th term is the 3rd term of 10th pair .

So it is

$\\ \sf\longmapsto \boxed{\bf \dfrac{3}{10}}$