Question

the percentage of natural numbers from 10 to 99 both inclusive which are the product of consecutive natural number is

Answer 1

Let the numbers from 51 to 100 form an Arithmetic Progression or for a simpler word, a sequence.

The first term is 51 and the common difference is 1.

Let 100 be the nthnth term.

⟹100=51+(n−1)1⟹100=51+(n−1)1

⟹100=51+n−1⟹100=51+n−1

⟹100=50+n⟹100=50+n

⟹n=100−50⟹n=100−50

⟹n=50⟹n=50

100100 is the 50th50th term

There are 50 terms in this A.P.

The formula for finding sum of n terms in an A.P. is :

Sn=n2[2a+(n−1)d]Sn=n2[2a+(n−1)d]_____________________________________(1)

where,

SnSn - Sum of n terms of an Arithmetic Progression.

n - Number of terms .

a - First term of the Arithmetic Progression.

d - Common Difference.

Putting values from the question in (1)

⟹S50=502[2(51)+(50−1)1]⟹S50=502[2(51)+(50−1)1]

⟹S50=25102+50⟹S50=25102+50

⟹S50=25152⟹S50=25152

⟹S50=3800⟹S50=3800

Therefore, sum of all natural numbers from 51 - 100 is 3800.

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Answer 2

Answer:

9

9---. is correct answer for this question

7