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प्रश्न
Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.
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उत्तर
Solution:
Numbers which are divisible by both 2 and 5 are the numbers which are
divisible by 10.
Thus we need to find the number of natural numbers between 101 and 999 which are divisible by 10.
The first number between 101 and 999 which is divisible by 10 is 110 And the last number between 101 and 999 which is divisible by 10 is 990 Using the formula for arithmetic progression
where first term ( a ) = 110, last term ( Tn ) = 990 and difference (d) =10
Tn=a+(n-1)d
990=110+(n-1)d
880=(n-1)10
n-1=88
n=89
Hence there are 89 natural numbers between 101 and 999 which are divisible by both 2 and 5.
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संबंधित प्रश्न
Find the sum of the following APs:
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Find the sum of the first 51 terms of the A.P: whose second term is 2 and the fourth term is 8.
Find the sum 3 + 11 + 19 + ... + 803
The 4th term of an AP is zero. Prove that its 25th term is triple its 11th term.
If 4 times the 4th term of an A.P. is equal to 18 times its 18th term, then find its 22nd term.
Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.
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Q.15
The nth term of an Arithmetic Progression (A.P.) is given by the relation Tn = 6(7 – n)..
Find:
- its first term and common difference
- sum of its first 25 terms
