Arithmetic Progressions | Exercise 5.2

Question 14

How many multiples of 4 lie between 10 and 250?

Solution

Understand the Question

The multiples of $4$ lying between $10$ and $250$ form an Arithmetic Progression (AP) with common difference $d = 4$ .
The first multiple after $10$ is $12$ , and the last multiple before $250$ is $248$ .
To find the total number of multiples, we use the $n^{\text{th}}$ term formula of an AP: $l = a + (n - 1)d$ .

Step 1 · Identify the Arithmetic Progression

Finding the first and last multiples of $4$ between $10$ and $250$ :

First multiple after $10$ : $10 \div 4 = 2$ with remainder $2 \implies 4 \times 3 = 12$
Last multiple before $250$ : $250 \div 4 = 62$ with remainder $2 \implies 4 \times 62 = 248$

The sequence of multiples is $12, 16, 20, \dots, 248$ , which forms an AP where:

First term, $a = 12$
Common difference, $d = 4$
Last term, $l = 248$

Step 2 · Calculate the Number of Terms

Using the formula for the $n^{\text{th}}$ term of an AP $l = a + (n - 1)d$

Substitute $a = 12$ , $d = 4$ , and $l = 248$ $248 = 12 + (n - 1)4$

\begin{aligned} 248 - 12 &= (n - 1)4 \\ 236 &= (n - 1)4 \end{aligned}

\begin{aligned} \dfrac{236}{4} &= n - 1 \\[0.6em] 59 &= n - 1 \end{aligned}

\begin{aligned} n &= 59 + 1 \\ n &= 60 \end{aligned}

Answer

60

Common Mistakes

Boundary Inclusion Error: Including $250$ directly without verifying if it is a multiple of $4$ . Always find the remainder to determine the correct last term ( $248$ ).
Off-by-One Error: Forgetting to add $1$ at the final stage of solving for $n$ ( $n = 59 + 1 = 60$ ).

More questions in Exercise 5.2

Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and $a_n$ the $n^{\text{th}}$ term of the AP:

Choose the correct choice in the following and justify :

(i) 30th term of the AP: $10, 7, 4, \dots$ , is (A) $97$ (B) $77$ (C) $-77$ (D) $-87$

(ii) 11th term of the AP: $-3, -\dfrac{1}{2}, 2, \dots$ , is (A) $28$ (B) $22$ (C) $-38$ (D) $-48\dfrac{1}{2}$

In the following APs, find the missing terms in the boxes :

Which term of the AP: $3, 8, 13, 18, \dots$ , is $78$ ?

Find the number of terms in each of the following APs:

(i) $7, 13, 19, \dots, 205$

(ii) $18, 15\dfrac{1}{2}, 13, \dots, -47$

Check whether $-150$ is a term of the AP: $11, 8, 5, 2, \dots$

Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

If the 3rd and the 9th terms of an AP are $4$ and $-8$ respectively, which term of this AP is zero?

Q10

The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

Q11

Which term of the AP : $3, 15, 27, 39, \dots$ will be $132$ more than its $54^{\text{th}}$ term?

Q12

Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Q13

How many three-digit numbers are divisible by 7?

Q14

How many multiples of 4 lie between 10 and 250?

Q15

For what value of $n$ , are the $n^{\text{th}}$ terms of two APs: $63, 65, 67, \dots$ and $3, 10, 17, \dots$ equal?

Q16

Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Q17

Find the 20th term from the last term of the AP : $3, 8, 13, \dots, 253$ .

Q18

The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Q19

Subba Rao started work in 1995 at an annual salary of ₹ 5000 and received an increment of ₹ 200 each year. In which year did his income reach ₹ 7000?

Q20

Ramkali saved ₹ 5 in the first week of a year and then increased her weekly savings by ₹ 1.75. If in the $n$ th week, her weekly savings become ₹ 20.75, find $n$ .