Ex.5.2 Q14 Arithmetic Progressions Solution - NCERT Maths Class 10

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Question

How many multiples of \(4\) lie between \(10\) and \(250\)?

 Video Solution
Arithmetic Progressions
Ex 5.2 | Question 14

Text Solution

What is Known?

Numbers between \(10\) and \(250\)

What is Unknown?

Multiples of \(4\) between \(10\) and \(250.\)

Reasoning:

\({a_n} = a + \left( {n - 1} \right)d\) is the general term of AP. Where \({a_n}\) is the \(n\rm{th}\) term, \(a\) is the first term, \(d\) is the common difference and \(n\) is the number of terms.

Steps: 

By Observation, First multiple of \(4\) that is greater than \(10\) is \(12.\)

Next will be \(16.\)

Therefore, the series will be as follows: \(12,\, 16,\, 20,\, 24,\, \dots\)

All these are divisible by \(4\) and thus, all these are terms of an A.P. with first term as \(12\) and common difference as \(4.\)

When we divide \(250\) by \(4,\) the remainder will be \(2.\) Therefore, \(250 – 2 = 248\) is divisible by \(4\) which is the largest multiple of \(4\) within \(250.\)

Hence the final series is as follows:

\[12,\, 16,\, 20,\, 24,\, \dots, 248\]

Let \(248\) be the \(n\rm{th}\) term of this A.P.

We know that the \(n\rm{th}\) term of an A.P. Series,

\[\begin{align}a &= 12\\d &= 4\\{a_n} &= 248\\{a_n} &= a + (n - 1)d\\248 &= 12 + (n - 1)4\\\frac{{236}}{4} &= n - 1\\n& = 60\end{align}\]

Therefore, there are \(60\) multiples of \(4\) between \(10\) and \(250.\)