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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
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Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd and 4th term is 14.
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If k, 2k − 1 and 2k + 1 are three consecutive terms of an A.P., the value of k is
Q.1
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In an A.P., the sum of its first n terms is 6n – n². Find is 25th term.
Find the sum of natural numbers between 1 to 140, which are divisible by 4.
Activity: Natural numbers between 1 to 140 divisible by 4 are, 4, 8, 12, 16,......, 136
Here d = 4, therefore this sequence is an A.P.
a = 4, d = 4, tn = 136, Sn = ?
tn = a + (n – 1)d
`square` = 4 + (n – 1) × 4
`square` = (n – 1) × 4
n = `square`
Now,
Sn = `"n"/2["a" + "t"_"n"]`
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Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.
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