Mathematics solution NCERT

Class 10 - Chapter 5: Arithmetic Progressions

NCERTChapter 5Solution- Exercise 5.1

1. An arithmetic progression is a list of numbers in which each term is obtained by
adding a fixed number to the preceding term except the first term.

This fixed number is called the common difference of the AP. Remember that it can be positive, negative or zero.

Ise ese samjho, man lo agar tumhare school ki fees abhi 15000 Rs hai or
every year school 2000 Rs increase karta hai to next year ki fees 17000 Rs hogi
and then 19000 Rs hogi.
yehi AP hai.

Is case me common difference hoga = 2000, first term = 15000, second term = 17000 and so on.....
AP = 15000, 17000, 19000 .........

AP ka standard form = a, a + d, a + 2d, a + 3d, . . . Here, a = first term and d = common diffference.
Agar hum school ki fees wale scenario me dekhte hain to ye standard form follow hoti hai.
a = 15000 and d = 2000, then AP form hogi if we put values in standard form.

Standard form is called the general form of an AP.

common difference = second term - first term

All Imp Formulae

1. General Form of AP

a, a + d, a + 2d, a + 3d, ...

Where,

a = First term
d = Common difference
n = Number of terms
a_n = nth term
S_n = Sum of n terms

2. Common Difference (d)

d = Second term − First term

d = a₂ − a₁

d = a₃ − a₂

3. nth Term Formula

a_n = a + (n − 1)d

4. Last Term Formula

l = a + (n − 1)d

Where l = last term

5. Finding Number of Terms (n)

a_n = a + (n − 1)d

n = [(a_n − a)/d] + 1

6. Sum of n Terms (First Formula)

S_n = n/2 [2a + (n − 1)d]

7. Sum of n Terms (Second Formula)

S_n = n/2 (a + l)

8. Relationship Between Terms

a_n = a_n-1 + d

9. Middle Term Formula (When Number of Terms is Odd)

Middle term position = (n + 1)/2

10. Important Short Tricks

First term = a
Second term = a + d
Third term = a + 2d
Fourth term = a + 3d
Fifth term = a + 4d

11. Most Important Formulas to Remember

nth Term:
a_n = a + (n − 1)d

Sum of n Terms:
S_n = n/2 [2a + (n − 1)d]

Alternative Sum Formula:
S_n = n/2 (a + l)

EXERCISE 5.1

Question 1 (i)

In which of the following situations does the list of numbers involved make an arithmetic progression, and why?

(i) The taxi fare after each km when the fare is ₹15 for the first km and ₹8 for each additional km.

Solution:

The taxi fare is

For the first kilometre = ₹15
For every additional kilometre = ₹8

Therefore, the total fare after each kilometre is

$$ 15,\;23,\;31,\;39,\;47,\ldots $$

Find the difference between consecutive terms.

$$ 23-15=8 $$ $$ 31-23=8 $$ $$ 39-31=8 $$ $$ 47-39=8 $$

The difference between every two consecutive terms is the same.

$$ d=8 $$

Conclusion:

Since the common difference is constant, the given list forms an Arithmetic Progression (AP).

Answer:

Yes, the list forms an Arithmetic Progression because the common difference is

$$ \boxed{8}. $$

Question 1 (ii)

In which of the following situations does the list of numbers involved make an arithmetic progression, and why?

(ii) The amount of air present in a cylinder when a vacuum pump removes ¼ of the air remaining in the cylinder at a time.

Solution:

Suppose the cylinder initially contains 100 units of air.

After each operation, one-fourth of the remaining air is removed.

Therefore, the amount of air left after each operation is

$$ 100,\;75,\;56.25,\;42.1875,\;\ldots $$

Find the differences between consecutive terms.

$$ 75-100=-25 $$ $$ 56.25-75=-18.75 $$ $$ 42.1875-56.25=-14.0625 $$

The differences are not equal.

$$ -25\ne-18.75\ne-14.0625 $$

Conclusion:

Since the difference between consecutive terms is not constant, the given list does not form an Arithmetic Progression (AP).

Answer:

No, the list does not form an Arithmetic Progression because the common difference is not constant.

Question 1 (iii)

In which of the following situations does the list of numbers involved make an arithmetic progression, and why?

(iii) The cost of digging a well after every metre of digging, when it costs ₹150 for the first metre and rises by ₹50 for each subsequent metre.

Solution:

The cost of digging is

For the first metre = ₹150
For every subsequent metre, the cost increases by ₹50.

Therefore, the costs for successive metres are

$$ 150,\;200,\;250,\;300,\;350,\;\ldots $$

Find the difference between consecutive terms.

$$ 200-150=50 $$ $$ 250-200=50 $$ $$ 300-250=50 $$ $$ 350-300=50 $$

The difference between every two consecutive terms is the same.

$$ d=50 $$

Conclusion:

Since the common difference is constant, the given list forms an Arithmetic Progression (AP).

Answer:

Yes, the list forms an Arithmetic Progression because the common difference is

$$ \boxed{50}. $$

Question 1 (iv)

In which of the following situations does the list of numbers involved make an arithmetic progression, and why?

(iv) The amount of money in the account every year, when ₹10000 is deposited at compound interest at 8% per annum.

Solution:

The initial amount deposited is

$$ ₹10000. $$

At 8% compound interest per annum, the amount after each year is

$$ 10000,\;10800,\;11664,\;12597.12,\;\ldots $$

Find the difference between consecutive terms.

$$ 10800-10000=800 $$ $$ 11664-10800=864 $$ $$ 12597.12-11664=933.12 $$

The differences are not equal.

$$ 800\ne864\ne933.12 $$

Conclusion:

Since the difference between consecutive terms is not constant, the given list does not form an Arithmetic Progression (AP).

In fact, each term is obtained by multiplying the previous term by

$$ 1.08. $$

Therefore, it forms a Geometric Progression (GP), not an Arithmetic Progression.

Answer:

No, the list does not form an Arithmetic Progression because the common difference is not constant.

Question 2 (i)

Write the first four terms of the Arithmetic Progression when the first term and the common difference are given.

Given:

$$ a=10,\qquad d=10 $$

Solution:

The first term is

$$ a=10. $$

The common difference is

$$ d=10. $$

Each successive term is obtained by adding the common difference to the previous term.

First term:

$$ a=10 $$

Second term:

$$ a+d=10+10=20 $$

Third term:

$$ a+2d=10+2(10)=30 $$

Fourth term:

$$ a+3d=10+3(10)=40 $$

Answer:

$$ \boxed{10,\;20,\;30,\;40} $$

Question 2 (ii)

Write the first four terms of the Arithmetic Progression when the first term and the common difference are given.

Given:

$$ a=-2,\qquad d=0 $$

Solution:

The first term is

$$ a=-2. $$

The common difference is

$$ d=0. $$

Since the common difference is zero, every term of the AP is the same.

First term:

$$ -2 $$

Second term:

$$ -2+0=-2 $$

Third term:

$$ -2+0=-2 $$

Fourth term:

$$ -2+0=-2 $$

Answer:

$$ \boxed{-2,\;-2,\;-2,\;-2} $$

Question 2 (iii)

Write the first four terms of the Arithmetic Progression when the first term and the common difference are given.

Given:

$$ a=4,\qquad d=-3 $$

Solution:

The first term is

$$ a=4. $$

The common difference is

$$ d=-3. $$

Each successive term is obtained by adding the common difference to the previous term.

First term:

$$ 4 $$

Second term:

$$ 4+(-3)=1 $$

Third term:

$$ 1+(-3)=-2 $$

Fourth term:

$$ -2+(-3)=-5 $$

Answer:

$$ \boxed{4,\;1,\;-2,\;-5} $$

Question 2 (iv)

Write the first four terms of the Arithmetic Progression when the first term and the common difference are given.

Given:

$$ a=-10,\qquad d=-2 $$

Solution:

The first term is

$$ a=-10. $$

The common difference is

$$ d=-2. $$

Each successive term is obtained by adding the common difference to the previous term.

First term:

$$ -10 $$

Second term:

$$ -10+(-2)=-12 $$

Third term:

$$ -12+(-2)=-14 $$

Fourth term:

$$ -14+(-2)=-16 $$

Answer:

$$ \boxed{-10,\;-12,\;-14,\;-16} $$

Question 2 (v)

Write the first four terms of the Arithmetic Progression when the first term and the common difference are given.

Given:

$$ a=-1,\qquad d=\frac12 $$

Solution:

The first term is

$$ a=-1. $$

The common difference is

$$ d=\frac12. $$

Each successive term is obtained by adding the common difference to the previous term.

First term:

$$ -1 $$

Second term:

$$ -1+\frac12=-\frac12 $$

Third term:

$$ -\frac12+\frac12=0 $$

Fourth term:

$$ 0+\frac12=\frac12 $$

Answer:

$$ \boxed{-1,\;-\frac12,\;0,\;\frac12} $$

Question 3 (i)

For the following Arithmetic Progressions (APs), write the first term and the common difference.

AP:

$$ 7,\;10,\;13,\;16,\;\ldots $$

Solution:

The first term of an AP is the first number in the sequence.

$$ a=7 $$

The common difference is obtained by subtracting the first term from the second term.

$$ d=10-7=3 $$

Verification:

$$ 13-10=3 $$ $$ 16-13=3 $$

Hence, the common difference is constant.

Answer:

$$ \boxed{a=7,\qquad d=3} $$

Question 3 (ii)

For the following Arithmetic Progressions (APs), write the first term and the common difference.

AP:

$$ -18,\;-15,\;-12,\;\ldots $$

Solution:

The first term is

$$ a=-18 $$

The common difference is

$$ d=-15-(-18) $$ $$ =3 $$

Verification:

$$ -12-(-15)=3 $$

Hence, the common difference is constant.

Answer:

$$ \boxed{a=-18,\qquad d=3} $$

Question 3 (iii)

For the following Arithmetic Progressions (APs), write the first term and the common difference.

AP:

$$ -5,\;-3,\;-1,\;1,\;\ldots $$

Solution:

The first term of the AP is

$$ a=-5 $$

The common difference is obtained by subtracting the first term from the second term.

$$ d=-3-(-5) $$ $$ =2 $$

Verification:

$$ -1-(-3)=2 $$ $$ 1-(-1)=2 $$

Since the difference between consecutive terms is the same, the common difference is

$$ d=2. $$

Answer:

$$ \boxed{a=-5,\qquad d=2} $$

Question 3 (iv)

For the following Arithmetic Progressions (APs), write the first term and the common difference.

AP:

$$ \frac12,\;1,\;\frac32,\;2,\;\ldots $$

Solution:

The first term is

$$ a=\frac12 $$

The common difference is

$$ d=1-\frac12 $$ $$ =\frac12 $$

Verification:

$$ \frac32-1=\frac12 $$ $$ 2-\frac32=\frac12 $$

Hence, the common difference is constant.

Answer:

$$ \boxed{a=\frac12,\qquad d=\frac12} $$

Question 4 (i)

Which of the following are APs? If they form an AP, find the common difference d and write three more terms.

(i)

$$ 2,\;4,\;8,\;16,\;\ldots $$

Solution:

Find the difference between consecutive terms.

$$ 4-2=2 $$ $$ 8-4=4 $$ $$ 16-8=8 $$

Since

$$ 2\ne4\ne8, $$

the common difference is not constant.

Answer:

The given sequence is not an Arithmetic Progression (AP).

Question 4 (ii)

(ii)

$$ 2,\;\frac{5}{2},\;3,\;\frac{7}{2},\;\ldots $$

Solution:

Find the difference between consecutive terms.

$$ \frac52-2=\frac12 $$ $$ 3-\frac52=\frac12 $$ $$ \frac72-3=\frac12 $$

The common difference is the same.

$$ d=\frac12 $$

Hence, the sequence is an Arithmetic Progression.

Add the common difference repeatedly to obtain the next three terms.

$$ \frac72+\frac12=4 $$ $$ 4+\frac12=\frac92 $$ $$ \frac92+\frac12=5 $$

Answer:

The given sequence is an Arithmetic Progression.

$$ \boxed{d=\frac12} $$

The next three terms are

$$ \boxed{4,\;\frac92,\;5} $$

Question 4 (iii)

(iii)

$$ -1.2,\;-3.2,\;-5.2,\;-7.2,\;\ldots $$

Solution:

Find the difference between consecutive terms.

$$ -3.2-(-1.2)=-2 $$ $$ -5.2-(-3.2)=-2 $$ $$ -7.2-(-5.2)=-2 $$

The common difference is the same.

$$ d=-2 $$

Hence, the sequence is an Arithmetic Progression.

Add the common difference repeatedly to obtain the next three terms.

$$ -7.2+(-2)=-9.2 $$ $$ -9.2+(-2)=-11.2 $$ $$ -11.2+(-2)=-13.2 $$

Answer:

The given sequence is an Arithmetic Progression.

$$ \boxed{d=-2} $$

The next three terms are

$$ \boxed{-9.2,\;-11.2,\;-13.2} $$

Question 4 (iv)

(iv)

$$ -10,\;-6,\;-2,\;2,\;\ldots $$

Solution:

Find the difference between consecutive terms.

$$ -6-(-10)=4 $$ $$ -2-(-6)=4 $$ $$ 2-(-2)=4 $$

The common difference is the same.

$$ d=4 $$

Hence, the sequence is an Arithmetic Progression.

Add the common difference repeatedly to obtain the next three terms.

$$ 2+4=6 $$ $$ 6+4=10 $$ $$ 10+4=14 $$

Answer:

The given sequence is an Arithmetic Progression.

$$ \boxed{d=4} $$

The next three terms are

$$ \boxed{6,\;10,\;14} $$

Question 4 (v)

(v)

$$ 3,\;3+\sqrt2,\;3+2\sqrt2,\;3+3\sqrt2,\;\ldots $$

Solution:

Find the difference between consecutive terms.

$$ (3+\sqrt2)-3=\sqrt2 $$ $$ (3+2\sqrt2)-(3+\sqrt2)=\sqrt2 $$ $$ (3+3\sqrt2)-(3+2\sqrt2)=\sqrt2 $$

The common difference is the same.

$$ d=\sqrt2 $$

Hence, the sequence is an Arithmetic Progression.

Add the common difference repeatedly to obtain the next three terms.

$$ 3+3\sqrt2+\sqrt2=3+4\sqrt2 $$ $$ 3+4\sqrt2+\sqrt2=3+5\sqrt2 $$ $$ 3+5\sqrt2+\sqrt2=3+6\sqrt2 $$

Answer:

The given sequence is an Arithmetic Progression.

$$ \boxed{d=\sqrt2} $$

The next three terms are

$$ \boxed{3+4\sqrt2,\;3+5\sqrt2,\;3+6\sqrt2} $$

Question 4 (vi)

(vi)

$$ 0.2,\;0.22,\;0.222,\;0.2222,\;\ldots $$

Solution:

Find the difference between consecutive terms.

$$ 0.22-0.2=0.02 $$ $$ 0.222-0.22=0.002 $$ $$ 0.2222-0.222=0.0002 $$

Since

$$ 0.02\ne0.002\ne0.0002, $$

the common difference is not constant.

Answer:

The given sequence is not an Arithmetic Progression (AP).

Question 4 (vii)

(vii)

$$ 0,\;-4,\;-8,\;-12,\;\ldots $$

Solution:

Find the difference between consecutive terms.

$$ -4-0=-4 $$ $$ -8-(-4)=-4 $$ $$ -12-(-8)=-4 $$

The common difference is the same.

$$ d=-4 $$

Hence, the sequence is an Arithmetic Progression.

Add the common difference repeatedly to obtain the next three terms.

$$ -12+(-4)=-16 $$ $$ -16+(-4)=-20 $$ $$ -20+(-4)=-24 $$

Answer:

The given sequence is an Arithmetic Progression.

$$ \boxed{d=-4} $$

The next three terms are

$$ \boxed{-16,\;-20,\;-24} $$

Question 4 (viii)

(viii)

$$ -\frac12,\;-\frac12,\;-\frac12,\;-\frac12,\;\ldots $$

Solution:

Find the difference between consecutive terms.

$$ -\frac12-\left(-\frac12\right)=0 $$ $$ -\frac12-\left(-\frac12\right)=0 $$ $$ -\frac12-\left(-\frac12\right)=0 $$

The common difference is the same.

$$ d=0 $$

Hence, the sequence is an Arithmetic Progression.

Add the common difference repeatedly to obtain the next three terms.

$$ -\frac12 $$ $$ -\frac12 $$ $$ -\frac12 $$

Answer:

The given sequence is an Arithmetic Progression.

$$ \boxed{d=0} $$

The next three terms are

$$ \boxed{-\frac12,\;-\frac12,\;-\frac12} $$

Question 4 (ix)

(ix)

$$ 1,\;3,\;9,\;27,\;\ldots $$

Solution:

Find the difference between consecutive terms.

$$ 3-1=2 $$ $$ 9-3=6 $$ $$ 27-9=18 $$

Since

$$ 2\ne6\ne18, $$

the common difference is not constant.

Answer:

The given sequence is not an Arithmetic Progression (AP).

Question 4 (x)

(x)

$$ a,\;2a,\;3a,\;4a,\;\ldots $$

Solution:

Find the difference between consecutive terms.

$$ 2a-a=a $$ $$ 3a-2a=a $$ $$ 4a-3a=a $$

The common difference is the same.

$$ d=a $$

Hence, the sequence is an Arithmetic Progression.

Add the common difference repeatedly to obtain the next three terms.

$$ 4a+a=5a $$ $$ 5a+a=6a $$ $$ 6a+a=7a $$

Answer:

The given sequence is an Arithmetic Progression.

$$ \boxed{d=a} $$

The next three terms are

$$ \boxed{5a,\;6a,\;7a} $$

Question 4 (xi)

(xi)

$$ a,\;a^2,\;a^3,\;a^4,\;\ldots $$

Solution:

Find the difference between consecutive terms.

$$ a^2-a $$ $$ a^3-a^2 $$ $$ a^4-a^3 $$

These differences are not equal in general.

Therefore, the common difference is not constant.

Answer:

The given sequence is not an Arithmetic Progression (AP).

Question 4 (xii)

(xii)

$$ \sqrt2,\;\sqrt8,\;\sqrt{18},\;\sqrt{32},\;\ldots $$

Solution:

Simplify each term.

$$ \sqrt8=2\sqrt2 $$ $$ \sqrt{18}=3\sqrt2 $$ $$ \sqrt{32}=4\sqrt2 $$

Thus, the sequence becomes

$$ \sqrt2,\;2\sqrt2,\;3\sqrt2,\;4\sqrt2,\;\ldots $$

Find the difference between consecutive terms.

$$ 2\sqrt2-\sqrt2=\sqrt2 $$ $$ 3\sqrt2-2\sqrt2=\sqrt2 $$ $$ 4\sqrt2-3\sqrt2=\sqrt2 $$

The common difference is the same.

$$ d=\sqrt2 $$

Hence, the sequence is an Arithmetic Progression.

Add the common difference repeatedly to obtain the next three terms.

$$ 4\sqrt2+\sqrt2=5\sqrt2 $$ $$ 5\sqrt2+\sqrt2=6\sqrt2 $$ $$ 6\sqrt2+\sqrt2=7\sqrt2 $$

Answer:

The given sequence is an Arithmetic Progression.

$$ \boxed{d=\sqrt2} $$

The next three terms are

$$ \boxed{5\sqrt2,\;6\sqrt2,\;7\sqrt2} $$

Question 4 (xiii)

(xiii)

$$ 1^2,\;2^2,\;3^2,\;4^2,\;\ldots $$

Solution:

Evaluate the terms.

$$ 1,\;4,\;9,\;16,\;\ldots $$

Find the difference between consecutive terms.

$$ 4-1=3 $$ $$ 9-4=5 $$ $$ 16-9=7 $$

Since

$$ 3\ne5\ne7, $$

the common difference is not constant.

Answer:

The given sequence is not an Arithmetic Progression (AP).

Question 4 (xiv)

(xiv)

$$ 1,\;\frac12,\;\frac13,\;\frac14,\;\ldots $$

Solution:

Find the difference between consecutive terms.

$$ \frac12-1=-\frac12 $$ $$ \frac13-\frac12=-\frac16 $$ $$ \frac14-\frac13=-\frac1{12} $$

Since

$$ -\frac12\ne-\frac16\ne-\frac1{12}, $$

the common difference is not constant.

Answer:

The given sequence is not an Arithmetic Progression (AP).

Question 4 (xv)

(xv)

$$ 1,\;\sqrt2,\;\sqrt3,\;2,\;\ldots $$

Solution:

Find the difference between consecutive terms.

$$ \sqrt2-1 $$ $$ \sqrt3-\sqrt2 $$ $$ 2-\sqrt3 $$

These differences are not equal.

Therefore, the common difference is not constant.

Answer:

The given sequence is not an Arithmetic Progression (AP).