
How many numbers from 101 to 200 are divisible by 5?
Answer
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Hint: First write the given condition into an algebraic sequence by using the given condition. Then you can see that you get an arithmetic progression. By using the formula of \[{{n}^{th}}\] the term finds the value of n. So we get that there are n terms which are divisible by 5. The value of n is the required result of the question. The formula of the \[{{n}^{th}}\] term of an arithmetic progression with first term a, and common difference d is given by:
Complete step-by-step answer:
The first term in the sequence is 105.
By assuming the value of common difference as 5 we can write the sequence as:
105, 110, 115,……, 200.
If an arithmetic progression has a first term as ‘a’ and a common difference as ‘d’. Then the \[{{n}^{th}}\] term of the arithmetic progression is given by:
\[{{t}_{n}}=a+\left( n-1 \right)d\]
By using the given sequence, we can say it as:
a = 105, d = 5, \[{{t}_{n}}\]= 200.
By substituting the given values into the formula, we get it as:
200 = 105 + (n-1) 5
By multiplying 5 inside the bracket, we get the equation as:
200 = 105 + 5n – 5
By simplifying the right hand side, we get it in the form of:
200 = 100 + 5n
By subtracting 100 on both sides of equation we can write it as:
200 – 100 = 100 + 5n – 100
By simplifying the above equation, we can write it in the form of:
100 = 5n
By dividing with 5 on both sides, we get it in the form of:
\[\dfrac{100}{5}=\dfrac{5n}{5}\]
By cancelling the common terms, we get it in the form of:
n = 20.
So the value 200 is the \[{{20}^{th}}\] term of the sequence.
So, there are 20 numbers divisible by 5 in the range 101 to 200. Therefore, this is the required result in the given question.
Note: Alternative way is: \[200=5\times 40\] and \[100=5\times 20.\] So, there are 40 numbers from 1 to 200 and 20 numbers from 1 to 100. So, (40 – 20) = 20 numbers will be there from 101 to 200. Be careful it is given 101 and you must not consider 100. Some students in a hurry consider 100 also. It is given from 101 to 200. So, you must consider the term 200. If they say in between 101 and 200, then 200 is neglected but not in this case. So, remember this point.
Complete step-by-step answer:
The first term in the sequence is 105.
By assuming the value of common difference as 5 we can write the sequence as:
105, 110, 115,……, 200.
If an arithmetic progression has a first term as ‘a’ and a common difference as ‘d’. Then the \[{{n}^{th}}\] term of the arithmetic progression is given by:
\[{{t}_{n}}=a+\left( n-1 \right)d\]
By using the given sequence, we can say it as:
a = 105, d = 5, \[{{t}_{n}}\]= 200.
By substituting the given values into the formula, we get it as:
200 = 105 + (n-1) 5
By multiplying 5 inside the bracket, we get the equation as:
200 = 105 + 5n – 5
By simplifying the right hand side, we get it in the form of:
200 = 100 + 5n
By subtracting 100 on both sides of equation we can write it as:
200 – 100 = 100 + 5n – 100
By simplifying the above equation, we can write it in the form of:
100 = 5n
By dividing with 5 on both sides, we get it in the form of:
\[\dfrac{100}{5}=\dfrac{5n}{5}\]
By cancelling the common terms, we get it in the form of:
n = 20.
So the value 200 is the \[{{20}^{th}}\] term of the sequence.
So, there are 20 numbers divisible by 5 in the range 101 to 200. Therefore, this is the required result in the given question.
Note: Alternative way is: \[200=5\times 40\] and \[100=5\times 20.\] So, there are 40 numbers from 1 to 200 and 20 numbers from 1 to 100. So, (40 – 20) = 20 numbers will be there from 101 to 200. Be careful it is given 101 and you must not consider 100. Some students in a hurry consider 100 also. It is given from 101 to 200. So, you must consider the term 200. If they say in between 101 and 200, then 200 is neglected but not in this case. So, remember this point.
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