Please Wait... While Loading Full Video#### Class 10 Chapter 5 - ARITHMETIC PROGRESSIONS

♦ nth Term of an AP

● Let us consider the situation again, given in Section 5.1 in which Reena applied for a job and got selected. She has been offered the job with a starting monthly salary of Rs. 8000, with an annual increment of Rs. 500. What would be her monthly salary for the fifth year?

● To answer this, let us first see what her monthly salary for the second year would be.

● It would be Rs. (8000 + 500) = Rs. 8500. In the same way, we can find the monthly salary for the `3rd, 4th` and `5th` year by adding Rs. 500 to the salary of the previous year. So, the salary for the 3rd year = Rs. (8500 + 500)

`= Rs. (8000 + 500 + 500)`

`= Rs. (8000 + 2 × 500)`

`= Rs. [8000 + (3 – 1) × 500]` (for the 3rd year)

`= Rs. 9000`

Salary for the 4th year `= Rs. (9000 + 500)`

`= Rs. (8000 + 500 + 500 + 500)`

`= Rs. (8000 + 3 × 500)`

`= Rs. [8000 + (4 – 1) × 500]` (for the 4th year)

`= Rs. 9500`

Salary for the 5th year = Rs. (9500 + 500 )

`= Rs. (8000+500+500+500 + 500)`

`= Rs. (8000 + 4 × 500)`

`= Rs. [8000 + (5 – 1) × 500]` (for the 5th year)

`= Rs. 10000`

Observe that we are getting a list of numbers `8000, 8500, 9000, 9500, 10000, . . .`

● These numbers are in AP.

● Now, looking at the pattern formed above, can you find her monthly salary for the 6th year? The 15th year? And, assuming that she will still be working in the job, what about the monthly salary for the 25th year?

You would calculate this by adding Rs. 500 each time to the salary of the previous year to give the answer. Can we make this process shorter?

Let us see. You may have already got some idea from the way we have obtained the salaries above.

Salary for the 15th year

= Salary for the 14th year + Rs. 500

● To answer this, let us first see what her monthly salary for the second year would be.

● It would be Rs. (8000 + 500) = Rs. 8500. In the same way, we can find the monthly salary for the `3rd, 4th` and `5th` year by adding Rs. 500 to the salary of the previous year. So, the salary for the 3rd year = Rs. (8500 + 500)

`= Rs. (8000 + 500 + 500)`

`= Rs. (8000 + 2 × 500)`

`= Rs. [8000 + (3 – 1) × 500]` (for the 3rd year)

`= Rs. 9000`

Salary for the 4th year `= Rs. (9000 + 500)`

`= Rs. (8000 + 500 + 500 + 500)`

`= Rs. (8000 + 3 × 500)`

`= Rs. [8000 + (4 – 1) × 500]` (for the 4th year)

`= Rs. 9500`

Salary for the 5th year = Rs. (9500 + 500 )

`= Rs. (8000+500+500+500 + 500)`

`= Rs. (8000 + 4 × 500)`

`= Rs. [8000 + (5 – 1) × 500]` (for the 5th year)

`= Rs. 10000`

Observe that we are getting a list of numbers `8000, 8500, 9000, 9500, 10000, . . .`

● These numbers are in AP.

● Now, looking at the pattern formed above, can you find her monthly salary for the 6th year? The 15th year? And, assuming that she will still be working in the job, what about the monthly salary for the 25th year?

You would calculate this by adding Rs. 500 each time to the salary of the previous year to give the answer. Can we make this process shorter?

Let us see. You may have already got some idea from the way we have obtained the salaries above.

Salary for the 15th year

= Salary for the 14th year + Rs. 500

Q 3149180013

Find the 10th term of the AP : 2, 7, 12, . . .

Class 10 Chapter 5 Example 3

Class 10 Chapter 5 Example 3

Here, a = 2, d = 7 – 2 = 5 and n = 10.

We have `a_n = a + (n – 1) d`

So, `a_10 = 2 + (10 – 1) × 5 = 2 + 45 = 47`

Therefore, the 10th term of the given AP is 47.

Q 3159180014

Which term of the AP : 21, 18, 15, . . . is – 81? Also, is any term 0? Give

reason for your answer.

Class 10 Chapter 5 Example 4

reason for your answer.

Class 10 Chapter 5 Example 4

Here, a = 21, d = 18 – 21 = – 3 and an = – 81, and we have to find n.

As `a_n = a + ( n – 1) d`,

we have – 81 = 21 + (n – 1)(– 3)

– 81 = 24 – 3n

– 105 = – 3n

So, n = 35

Therefore, the 35th term of the given AP is – 81.

Next, we want to know if there is any n for which `a_n = 0`. If such an n is there, then

21 + (n – 1) (–3) = 0,

i.e., 3(n – 1) = 21

i.e., n = 8

So, the eighth term is 0.

Q 3169180015

Determine the AP whose 3rd term is 5 and the 7th term is 9.

Class 10 Chapter 5 Example 5

Class 10 Chapter 5 Example 5

We have

`a_3 = a + (3 – 1) d = a + 2d = 5` (1)

and `a_7 = a + (7 – 1) d = a + 6d = 9` (2)

Solving the pair of linear equations (1) and (2), we get

`a = 3, d = 1`

Hence, the required AP is 3, 4, 5, 6, 7, . . .

Q 3189180017

Check whether 301 is a term of the list of numbers 5, 11, 17, 23, . . .

Class 10 Chapter 5 Example 6

Class 10 Chapter 5 Example 6

We have :

`a_2 – a_1 = 11 – 5 = 6, a_3 – a_2 = 17 – 11 = 6, a_4 – a_3 = 23 – 17 = 6`

As `a_(k + 1) – a_k` is the same for k = 1, 2, 3, etc., the given list of numbers is an AP.

Now, a = 5 and d = 6.

Let 301 be a term, say, the nth term of the this AP.

We know that

`a_n = a + (n – 1) d`

So, `301 = 5 + (n – 1) × 6`

i.e., `301 = 6n – 1`

So, ` n = 302/6 =151/3`

But n should be a positive integer (Why?). So, 301 is not a term of the given list of

numbers.

Q 3109180018

How many two-digit numbers are divisible by 3?

Class 10 Chapter 5 Example 7

Class 10 Chapter 5 Example 7

The list of two-digit numbers divisible by 3 is :

12, 15, 18, . . . , 99

Is this an AP? Yes it is. Here, `a = 12, d = 3, a_n = 99`.

As `a_n = a + (n – 1) d`,

we have `99 = 12 + (n – 1) × 3`

i.e., 87 = (n – 1) × 3

i.e., n – 1 = 87/3 = 29

i.e., n = 29 + 1 = 30

So, there are 30 two-digit numbers divisible by 3.

Q 3119180019

Find the 11th term from the last term (towards the first term) of the

AP : 10, 7, 4, . . ., – 62.

Class 10 Chapter 5 Example 8

AP : 10, 7, 4, . . ., – 62.

Class 10 Chapter 5 Example 8

Here, a = 10, d = 7 – 10 = – 3, l = – 62,

where l = a + (n – 1) d

To find the 11th term from the last term, we will find the total number of terms in

the AP.

So, – 62 = 10 + (n – 1)(–3)

i.e., – 72 = (n – 1)(–3)

i.e., n – 1 = 24

or n = 25

So, there are 25 terms in the given AP.

The 11th term from the last term will be the 15th term. (Note that it will not be

the 14th term. Why?)

So, `a_15 = 10 + (15 – 1)(–3) = 10 – 42 = – 32`

i.e., the 11th term from the last term is – 32.

Alternative Solution :

If we write the given AP in the reverse order, then a = – 62 and d = 3 (Why?)

So, the question now becomes finding the 11th term with these a and d.

So, `a_11 = – 62 + (11 – 1) × 3 = – 62 + 30 = – 32`

So, the 11th term, which is now the required term, is – 32.

Q 3129280111

A sum of Rs. 1000 is invested at 8% simple interest per year. Calculate the

interest at the end of each year. Do these interests form an AP? If so, find the interest

at the end of 30 years making use of this fact.

Class 10 Chapter 5 Example 9

interest at the end of each year. Do these interests form an AP? If so, find the interest

at the end of 30 years making use of this fact.

Class 10 Chapter 5 Example 9

We know that the formula to calculate simple interest is given by

Simple Interest `= (P xx R xx T)/(100 )`

So, the interest at the end of the 1st year = Rs. `(1000 xx 8 xx 1)/(100)=` Rs . 80

The interest at the end of the 2nd year = Rs. ` (1000 xx 8 xx 2 )/100 = ` Rs. 160

The interest at the end of the 3rd year = Rs. ` (1000 xx 8 xx 3)/100 =` Rs . 240

Similarly, we can obtain the interest at the end of the 4th year, 5th year, and so on.

So, the interest (in Rs) at the end of the 1st, 2nd, 3rd, . . . years, respectively are

80, 160, 240, . . .

It is an AP as the difference between the consecutive terms in the list is 80, i.e.,

d = 80. Also, a = 80.

So, to find the interest at the end of 30 years, we shall find `a_30`.

Now, `a_30 = a + (30 – 1) d = 80 + 29 × 80 = 2400`

So, the interest at the end of 30 years will be Rs. 2400.

Q 3139280112

In a flower bed, there are 23 rose plants in the first row, 21 in the

second, 19 in the third, and so on. There are 5 rose plants in the last row. How many

rows are there in the flower bed?

Class 10 Chapter 5 Example 10

second, 19 in the third, and so on. There are 5 rose plants in the last row. How many

rows are there in the flower bed?

Class 10 Chapter 5 Example 10

The number of rose plants in the 1st, 2nd, 3rd, . . ., rows are :

23, 21, 19, . . ., 5

It forms an AP (Why?). Let the number of rows in the flower bed be n.

Then a = 23, d = 21 – 23 = – 2, `a_n = 5`

As, `a_n = a + (n – 1) d`

We have, 5 = 23 + (n – 1)(– 2)

i.e., – 18 = (n – 1)(– 2)

i.e., n = 10

So, there are 10 rows in the flower bed.