Question

Natural numbers starting from 10 are written in a row as given below:

10 11 12 13 14 15.............
what will be the 4000th digit in the sequence? ​

Answer 1

Step-by-step explanation:

Last digit will be 9

Step-by-step explanation:

Let try to figure out which number we will be writing when it is 4000th

digit

It is starting from 10 . So there will be 90 , 2 digit number (10 to 99)

Similarly 900 3 digit , 9000 4 digit.

Now you can see that will be a 4 digit number.

90 *2 + 900*3 = 2880 < 4000

When we will write last 3 digit number (999) it is 2880th digit.

So 4000 - 2880 = 1120 more to go.

So which 4 digit number we will be writing

k = \frac{1120}{4} = 280k=

4

1120

=280

So we have completed write

999 + 280 = 1279

And last digit 9 .

Answer 2

4000th term would be 4009

Step-by-step explanation:

Given sequence,

10 11 12 13 14 15.............,

Which is an AP,

Having first term, a = 10,

Common difference, d = 1,

Thus, the nth term of the sequence,

$a(n) = a + (n-1)d$

$=10+(n-1)$

$=9+n$

If n = 4000,

4000th term = 9 + 4000 = 4009

#Learn more:

Find the 15th term of the sequence 20,15,10​........

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