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Question
How many two-digit numbers are divisible by 5?
Activity :- Two-digit numbers divisible by 5 are, 10, 15, 20, ......, 95.
Here, d = 5, therefore this sequence is an A.P.
Here, a = 10, d = 5, tn = 95, n = ?
tn = a + (n − 1) `square`
`square` = 10 + (n – 1) × 5
`square` = (n – 1) × 5
`square` = (n – 1)
Therefore n = `square`
There are `square` two-digit numbers divisible by 5
Solution
Two-digit numbers divisible by 5 are, 10, 15, 20, ......, 95.
Here, d = 5, therefore this sequence is an A.P.
Here, a = 10, d = 5, tn = 95, n = ?
tn = a + (n − 1) d
∴ 95 = 10 + (n – 1) × 5
∴ 95 – 10 = (n – 1) × 5
∴ 85 = (n – 1) × 5
∴ `85/5` = (n – 1)
∴ 17 = (n – 1)
∴ n = 17 + 1
Therefore n = 18
There are 18 two-digit numbers divisible by 5
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Activity :- Here, d = `square`, therefore this sequence is an A.P.
a = 5, d = `square`
Let nth term of this A.P. be 301
tn = a + (n – 1) `square`
301 = 5 + (n – 1) × `square`
301 = 6n – 1
n = `302/6` = `square`
But n is not positive integer.
Therefore, 301 is `square` the term of sequence 5, 11, 17, 23, ......
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