Arithmetic Progressions ⇒⇒ 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 Rs 15 for the first km and Rs 8 for each additional km

Solution :

According to the question

For the first kilometer, the fare is ₹15 .

For the second kilometer, the fare is ₹15 + ₹8 = ₹23

For the third kilometer, the fare is ₹23 + ₹8 = ₹31.

For the fourth kilometer, the fare is ₹31 + ₹8 = ₹39.

Since the difference between each consecutive term is a constant 8, the list of numbers forms an arithmetic progression.

In this case, the first term $a_1$ is 15 and the common difference (d) is 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 $ 1 \over 4 $ of the air remaining in the cylinder at a time .

Solution :

According to the question

Let the initial volume of air in a cylinder be $ a_1 $ = V

Amount of air present in the cylinder after first-time removal $ 1 \over 4 $ of air at a time

$$ a_2 = { V - {1 \over 4}V } = {3 \over 4 }V $$

Amount of air present in the cylinder after second-time removal

$$ a_3 = {{3 \over 4 }V - ({ 1 \over 4 } × { 3 \over 4}) V }$$

$$ a_3 = {{9 \over 16}V }$$

∴ $ a_1 $, $ a_2 $,$ a_3 $ . . . . ...

$ V, {3 \over 4 }V, {{9 \over 16}V }$. . ..

To check if it's an AP, we calculate the difference between consecutive terms:

Here difference between 2nd and 1st term

$$ a_2 - a_1 = {{3 \over 4 }V } - V $$

$$ ⇒{ a_2 - a_1 = {- V \over 4 } }$$

Here difference between 3rd and 2nd term

$$ a_3 - a_2 = {{9 \over 16}V }-{{3 \over 4 }V } $$

$$ ⇒{ a_3 - a_2 = {- 3V \over 16 } }$$

$$ { a_2 - a_1 } \ne {a_3 - a_2 } $$

Since the difference is not constant, this is not an arithmetic progression.
∴ a, a2, a3 does not form an A.P (AP)

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 meter of digging, when it costs Rs 150 for the first meter and rises by Rs 50 for each subsequent meter.

Solution :

According to the question

Cost for the first metre: = $ a_1 $ = ₹150 .

Cost for the second metre: = $ a_2 $ = ₹150 + ₹50 = ₹200

Cost for the third metre: = $ a_3 $ = ₹200 + ₹50 = ₹250.

Cost for the fourth metre: = $ a_4 $ = ₹250 + ₹50 = ₹300

List of number formed after above process

${ 150, 200, 250 }$. . ..

To check if it's an AP, we calculate the difference between consecutive terms:

Here difference between 2nd and 1st term

$$ a_2 - a_1 = 200 - 150 = 50 $$

Here difference between 3rd and 2nd term

$$ a_3 - a_2 = 250 - 200 = 50 $$

$$ { a_2 - a_1 } = {a_3 - a_2 } $$

The difference is a constant ₹50. Since the difference between consecutive terms is fixed, the list of numbers forms an arithmetic progression.

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 :

According to the question

We know that amount of present value P at r% compound interest after ‘n’ years.

$$ A = { [ P ( 1 + {r \over 100 })]^n } $$

Therefore, amount at the end of 1st year will be

$$ { [ 10000 ( 1 + {8 \over 100 })]^1 } = 10800 $$

amount at the end of 2nd year will be

$$ { [ 10000 ( 1 + {8 \over 100 })]^2 } = 11664 $$

amount at the end of 3rd year will be

$$ { [ 10000 ( 1 + {8 \over 100 })]^3 } = 12,597.12$$

List of amounts after above process

${ 10000, 10800, 11664 }$. . ..

Here difference between 2nd and 1st term

$$ a_2 - a_1 = 10800 - 10000 = 800 $$

Here difference between 3rd and 2nd term

$$ a_3 - a_2 = 11664 - 10800 = 864 $$

We can clearly see that.

$$ { a_2 - a_1 } \ne {a_3 - a_2 } $$

Since the difference between consecutive terms (864 and 800) is not constant, the sequence of amounts does not form an arithmetic progression.

Question 2 (i)

Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(i) a = 10, d = 10

Solution :

We know that

$$ a_n = { a + ( n - 1) d } $$

First term = a = 10

Therefore,second term will be

$$ { 10 + 1(10) } = 20 $$

third term will be

$$ { 10 + 2(10) } = 30 $$

fourth term will be

$$ { 10 + 3(10) } = 40 $$

The first four terms of AP are

${ 10, 20, 30,40, }$

Question 2 (ii)

Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(ii) a = -2, d = 0

Solution :

We know that

$$ a_n = { a + ( n - 1) d } $$

First term = a = -2

Therefore,second term will be

$$ { -2 + 1(0) } = -2 $$

third term will be

$$ { -2 + 2(0) } = -2 $$

fourth term will be

$$ { -2 + 3(0) } = -2 $$

The first four terms of AP are

${-2,-2,-2,-2 }$

Question 2 (iii)

Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(iii) a = 4, d = -3

Solution :

We know that

$$ a_n = { a + ( n - 1) d } $$

First term = a = 4

Therefore,second term will be

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

third term will be

$$ { 4 + 2(-3) } = -2 $$

fourth term will be

$$ { 4 + 3(-3) } = -5 $$

The first four terms of AP are

${ 4 , 1 , -2 , -5 }$

Question 2 (iv)

Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(iv) a = – 1 , d = 1/2

Solution :

We know that

$$ a_n = { a + ( n - 1) d } $$

First term = a = -1

Therefore,second term will be

$$ { -1 + 1 ({1 \over 2} ) } = {-1 \over 2} $$

third term will be

$$ { -1 + 2 ({1 \over 2} ) } = 0 $$

fourth term will be

$$ { -1 + 3 ({1 \over 2} ) } = {1 \over 2} $$

The first four terms of AP are

${ -1 , {-1 \over 2} , 0 , {1 \over 2} }$

Question 2 (v)

Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(v) a = – 1.25, d = – 0.25

Solution :

We know that

$$ a_n = { a + ( n - 1) d } $$

First term = a = – 1.25

Therefore,second term will be

$$ { – 1.25 + 1(-0.25 ) } = –1.5 $$

third term will be

$$ { – 1.25 + 2(-0.25 ) } = –1.75$$

fourth term will be

$$ { – 1.25 + 3(-0.25 ) } = –2 $$

The first four terms of AP are

${ – 1.25 , – 1.5 , – 1.75 , -2 }$

Question 3 (i)

For the following APs, write the first term and the common difference:
(i) 3, 1, – 1, – 3, ...

Solution :

The AP is 3, 1, – 1, – 3, …

∴ First term = $ { a_1 } $ = 3

Common difference d = $ { a_2 - a_1 } $

1 - 3 = - 2

First term is 3 and the common difference is -2 .

Question 3 (ii)

For the following APs, write the first term and the common difference:
(ii) -5, -1, 3, 7....

Solution :

The AP is -5, -1, 3, 7, …

∴ First term = $ { a_1 } $ = -5

Common difference d = $ { a_2 - a_1 } $

-1 - (-5) = 4

First term is -5 and the common difference is 4 .

Question 3 (iii)

For the following APs, write the first term and the common difference:
(iii) ${1\over 3} , {5\over 3}, {9\over 3},{13\over 3}, … $

Solution :

The AP is ${1\over 3} , {5\over 3}, {9\over 3},{13\over 3},… $

∴ First term = $ { a_1 } = {1\over 3} $

Common difference d = $ { a_2 - a_1 } $

$$ ⇒{5\over 3} - {1\over 3} $$

$$ ⇒{4\over 3} $$

First term is ${1\over 3}$ and the common difference is $ {4\over 3} $ .

Question 3 (iv)

For the following APs, write the first term and the common difference:
(iv) 0.6, 1.7, 2.8, 3.9,….

Solution :

The AP is 0.6, 1.7, 2.8, 3.9, …

∴ First term = 0.6

Common difference d = $ { a_2 - a_1 } $

1.7 - 0.6 = 1.1

First term is 0.6 and the common difference is 1.1.

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, …...

Solution :

The list of numbers is 2, 4, 8, 16, …

First term = $ { a_1 } $ = 2

Second term = $ { a_2 } $ = 4

Third term = $ { a_3 } $ = 8

Common difference d = $$ { a_2 - a_1 } = { 4 - 2 = 2 }$$

$$ { a_3 - a_2 } = { 8 - 4 = 4 }$$

We can clearly see that the difference between terms are not equal

$$ { a_3 - a_2 } \ne { a_2 - a_1 }$$

Thus given list of numbers do not form an Arithmetic Progression (AP)

Question 4 (ii)

Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
(ii) 2, $ { 5 \over 2 } $, 3,$ { 7 \over 2 } $, …...

Solution :

The list of numbers is 2, $ { 5 \over 2 } $, 3,$ { 7 \over 2 } $, …

First term = $ { a_1 } $ = 2

Second term = $ { a_2 } = { 5 \over 2 }$

Third term = $ { a_3 } $ = 3

Common difference d = $$ { a_2 - a_1 } = {{ 5 \over 2 } - 2 = { 1 \over 2 } }$$

$$ { a_3 - a_2 } = { 3 - { 5 \over 2 } = {1 \over 2 } }$$

We can clearly see that the difference between terms are equal and equal to $ { 1 \over 2 } $

$$ { a_3 - a_2 } = { a_2 - a_1 }$$

Thus given list of numbers form an Arithmetic Progression (AP)

We know that

$$ a_n = { a + ( n - 1) d } $$

∴ Next three terms :

$ { a_5 } = { 2 +(5-1) } × { 1 \over 2 } = 4 $

$ { a_6 } = { 2 +(6-1) } × { 1 \over 2 } = { 9 \over 2 } $

$ { a_7 } = { 2 +(7-1) } × { 1 \over 2 } = 5 $

Question 4 (iii)

Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
(iii) – 1.2, – 3.2, – 5.2, – 7.2....

Solution :

The list of numbers is – 1.2, – 3.2, – 5.2, – 7.2…

First term = $ { a_1 } $ = -1.2

Second term = $ { a_2 } = -3.2 $

Third term = $ { a_3 } $ = -5.2

Common difference d = $$ { a_2 - a_1 } = -3.2 - (-1.2) = -2 $$

$$ { a_3 - a_2 } = = -5.2 - (-3.2) = -2 $$

We can clearly see that the difference between terms are equal and equal to $ -2 $

$$ { a_3 - a_2 } = { a_2 - a_1 }$$

Thus given list of numbers form an Arithmetic Progression (AP)

We know that

$$ a_n = { a + ( n - 1) d } $$

∴ Next three terms :

$ { a_5 } = { – 1.2 +(5-1) } × (-2) = – 9.2 $

$ { a_6 } = { – 1.2 +(6-1) } × (-2) = – 11.2 $

$ { a_7 } = { – 1.2 +(7-1) } × (-2) = – 13.2 $

Question 4 (iv)

Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
(iv) – 10, – 6, – 2, 2, ...

Solution :

The list of numbers is – 10, – 6, – 2, 2, . ...

First term = $ { a_1 } $ = -10

Second term = $ { a_2 } = -6 $

Third term = $ { a_3 } $ = -2

Common difference d = $$ { a_2 - a_1 } = -6 - (-10) = 4 $$

$$ { a_3 - a_2 } = = - 2 - (-6) = 4 $$

We can clearly see that the difference between terms are equal and equal to $ 4 $

$$ { a_3 - a_2 } = { a_2 - a_1 }$$

Thus given list of numbers form an Arithmetic Progression (AP)

We know that

$$ a_n = { a + ( n - 1) d } $$

∴ Next three terms :

$ { a_5 } = { – 10 +(5-1) } × (4) = 6 $

$ { a_6 } = { – 10 +(6-1) } ×(4) = 10 $

$ { a_7 } = { – 10 +(7-1) } × (4) = 14 $

Question 4 (v)

Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
(v) 3,$( {3+\sqrt 2}) $, $ ({3+2\sqrt 2} )$, $( {3+3\sqrt 2}) $, ...

Solution :

The list of numbers is 3,$( {3+\sqrt 2}) $, $ ({3+2\sqrt 2} )$, $( {3+3\sqrt 2}) $, .....

First term = $ { a_1 } $ = 3

Second term = $ { a_2 } = ( {3+\sqrt 2}) $

Third term = $ { a_3 } = ( {3+2\sqrt 2})$

Common difference d = $$ { a_2 - a_1 } = ( {3+\sqrt 2}) - 3 $$

$$ = \sqrt 2 $$

$$ { a_3 - a_2 } = ({3+2\sqrt 2} ) - ({3+\sqrt 2} ) $$

$$ = ({3+2\sqrt 2} - 3 -{\sqrt 2} ) $$

$$ = \sqrt 2 $$

$$ { a_4 - a_3 } = ({3+3\sqrt 2} ) - ({3+2\sqrt 2} ) $$

$$ = ({3+3\sqrt 2} - 3 -{2\sqrt 2} ) $$

$$ = \sqrt 2 $$

We can clearly see that the difference between terms are equal and equal to $ \sqrt 2 $

$$ { a_3 - a_2 } = { a_2 - a_1 }$$

Thus given list of numbers form an Arithmetic Progression (AP)

We know that

$$ a_n = { a + ( n - 1) d } $$

∴ Next three terms :

$ { a_5 } = { 3 +(5-1) } × ( \sqrt 2 ) = ({3+4\sqrt 2} ) $

$ { a_6 } = { 3 +(6-1) } × ( \sqrt 2 ) = ({3+5\sqrt 2} ) $

$ { a_7 } = { 3 +(7-1) } × ( \sqrt 2 ) = ({3+6\sqrt 2} ) $

Question 4 (vi)

Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
(vi) 0.2, 0.22, 0.222, 0.2222,….

Solution :

The list of numbers is 0.2, 0.22, 0.222, 0.2222,…

First term = $ { a_1 } $ = 0.2

Second term = $ { a_2 } $ = 0.22

Third term = $ { a_3 } $ = 0.222

Common difference d = $$ { a_2 - a_1 } = { 0.22 - 0.2 = 0.02 }$$

$$ { a_3 - a_2 } = { 0.222 - 0.22 = 0.002 }$$

We can clearly see that the difference between terms are not equal

$$ { a_3 - a_2 } \ne { a_2 - a_1 }$$

Thus given list of numbers do not form an Arithmetic Progression (AP)

Question 4 (vii)

Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
(vii) 0, – 4, – 8, – 12,….

Solution :

The list of numbers is 0, – 4, – 8, – 12,…..

First term = $ { a_1 } $ = 0

Second term = $ { a_2 } = -4 $

Third term = $ { a_3 } $ = -8

Common difference d = $$ { a_2 - a_1 } = -4 - 0 = -4 $$

$$ { a_3 - a_2 } = - 8 - (-4) = -4 $$

We can clearly see that the difference between terms are equal and equal to $ - 4 $

$$ { a_3 - a_2 } = { a_2 - a_1 }$$

Thus given list of numbers form an Arithmetic Progression (AP)

We know that

$$ a_n = { a + ( n - 1) d } $$

∴ Next three terms :

$ { a_5 } = { 0 +(5-1) }× (-4) = -16 $

$ { a_6 } = { 0 +(6-1) } × (-4) = -20 $

$ { a_7 } = { 0 +(7-1) } × (-4) = -24 $

Question 4 (viii)

Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
(viii) $ -{ 1 \over 2 } $, $ -{ 1 \over 2 } $, $ -{ 1 \over 2 } $,$ -{ 1 \over 2 } $, …...

Solution :

The list of numbers is $ -{ 1 \over 2 } $, $ -{ 1 \over 2 } $, $ -{ 1 \over 2 } $,$ -{ 1 \over 2 } $, …...

First term = $ { a_1 } $ = $ -{ 1 \over 2 } $

Second term = $ { a_2 } $ = $ -{ 1 \over 2 } $

Third term = $ { a_3 } $ = $ -{ 1 \over 2 } $

Common difference d = $$ { a_2 - a_1 } = -{ 1 \over 2 } - (-{ 1 \over 2 }) = 0 $$

$$ { a_3 - a_2 } = -{ 1 \over 2 } - (-{ 1 \over 2 }) = 0 $$

We can clearly see that the difference between terms are equal and equal to 0

$$ { a_3 - a_2 } = { a_2 - a_1 }$$

Thus given list of numbers form an Arithmetic Progression (AP)

We know that

$$ a_n = { a + ( n - 1) d } $$

∴ Next three terms :

$ { a_5 } = { -{ 1 \over 2 } +(5-1) } × {0} = -{ 1 \over 2 } $

$ { a_6 } = { -{ 1 \over 2 } +(6-1) } × { 0 } = -{ 1 \over 2 } $

$ { a_7 } = { -{ 1 \over 2 } +(7-1) } × { 0 } = -{ 1 \over 2 } $

Question 4 (ix)

Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
(ix) 1, 3, 9, 27,….

Solution :

The list of numbers is 1, 3, 9, 27,……

First term = $ { a_1 } $ = 1

Second term = $ { a_2 } $ = 3

Third term = $ { a_3 } $ = 9

Common difference d = $$ { a_2 - a_1 } = { 3 - 1 = 2 }$$

$$ { a_3 - a_2 } = { 9 - 3 = 6 }$$

We can clearly see that the difference between terms are not equal

$$ { a_3 - a_2 } \ne { a_2 - a_1 }$$

Thus given list of numbers do not form an Arithmetic Progression (AP)

Question 4 (x)

Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
(x) a, 2a, 3a, 4a,….

Solution :

The list of numbers is a, 2a, 3a, 4a,…..

First term = $ { a_1 } $ = a

Second term = $ { a_2 } = 2a $

Third term = $ { a_3 } $ = 3a

Common difference d = $$ { a_2 - a_1 } = 2a - a = a $$

$$ { a_3 - a_2 } = 3a - 2a = a $$

We can clearly see that the difference between terms are equal and equal to $a $

$$ { a_3 - a_2 } = { a_2 - a_1 }$$

Thus given list of numbers form an Arithmetic Progression (AP)

We know that

$$ a_n = { a + ( n - 1) d } $$

∴ Next three terms :

$ { a_5 } = { a +(5-1) } × (a) = 5a $

$ { a_6 } = { a +(6-1) } × (a) = 6a $

$ { a_7 } = { a+(7-1) } × (a) = 7a $

Question 4 (xi)

Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
(xi) a, $ {a^2} $, $ {a^3} $, $ {a^4} $,….….

Solution :

The list of numbers is a, $ {a^2} $, $ {a^3} $, $ {a^4} $,….….

First term = $ { a_1 } $ = a

Second term = $ { a_2 } $ = $ {a^2} $

Third term = $ { a_3 } $ = $ {a^3} $

Common difference d = $$ { a_2 - a_1 } = {a^2} -a = a (a – 1) $$

$$ { a_3 - a_2 } = {a^3} - {a^2} = {a^2} (a – 1) $$

We can clearly see that the difference between terms are not equal

$$ { a_3 - a_2 } \ne { a_2 - a_1 }$$

Thus given list of numbers do not form an Arithmetic Progression (AP)

Question 4 (xii)

Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
(xii) $ {\sqrt 2} $, $ {\sqrt 8} $, $ {\sqrt 18} $, $ {\sqrt 32}$, ...

Solution :

The list of numbers is $ {\sqrt 2} $, $ {\sqrt 8} $, $ {\sqrt 18} $, $ {\sqrt 32}$, ...

First term = $ { a_1 } $ = $ {\sqrt 2} $

Second term = $ { a_2 } $ = $ {\sqrt 8} $

Third term = $ { a_3 } = $ $ {\sqrt 18} $

Common difference d = $$ { a_2 - a_1 } = {\sqrt 8} - {\sqrt 2} $$

$$ = {2\sqrt 2} - {\sqrt 2} $$

$$ = {\sqrt 2} $$

$$ { a_3 - a_2 } = {\sqrt 18} - {\sqrt 8} $$

$$ = {3\sqrt 2} - {2\sqrt 2} $$

$$ = \sqrt 2 $$

$$ { a_4 - a_3 } = {\sqrt 32} - {\sqrt 18} $$

$$ = {4\sqrt 2} - {3\sqrt 2} $$

$$ = \sqrt 2 $$

We can clearly see that the difference between terms are equal and equal to $ \sqrt 2 $

$$ { a_3 - a_2 } = { a_2 - a_1 }$$

Thus given list of numbers form an Arithmetic Progression (AP)

We know that

$$ a_n = { a + ( n - 1) d } $$

∴ Next three terms :

$ { a_5 } = { {\sqrt 2} +(5-1) } × ( \sqrt 2 ) = 5( \sqrt 2 ) = \sqrt 50 $

$ { a_6 } = { {\sqrt 2} +(6-1) } × ( \sqrt 2 ) = 6( \sqrt 2 ) = \sqrt 72 $

$ { a_7 } = { {\sqrt 2} +(7-1) } × ( \sqrt 2 ) = 7( \sqrt 2 ) = \sqrt 98 $

Question 4 (xiii)

Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
(xiii) $ {\sqrt 3} $, $ {\sqrt 6} $, $ {\sqrt 9} $, $ {\sqrt 12}$, ..

Solution :

The list of numbers is $ {\sqrt 3} $, $ {\sqrt 6} $, $ {\sqrt 9} $, $ {\sqrt 12}$, ..

First term = $ { a_1 } $ = ${\sqrt 3} $

Second term = $ { a_2 } $ = $ {\sqrt 6} $

Third term = $ { a_3 } = $ $ {\sqrt 9} $

Common difference d = $$ { a_2 - a_1 } = {\sqrt 6} - {\sqrt 3} $$

$$ = {\sqrt 3}({\sqrt 2} - 1 ) $$

$$ { a_3 - a_2 } = {\sqrt 9} - {\sqrt 6} $$

$$ = {\sqrt 3}({\sqrt 3} - {\sqrt 2} ) $$

$$ { a_4 - a_3 } = {\sqrt 12} - {\sqrt 9} $$

$$ = {\sqrt 3}({\sqrt 4} - {\sqrt 3} ) $$

We can clearly see that the difference between terms are not equal

$$ { a_3 - a_2 } \ne { a_2 - a_1 }$$

Thus given list of numbers do not form an Arithmetic Progression (AP)

Question 4 (xiv)

Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
(xiv) $ {1^2} $, $ {3^2} $, $ {5^2} $, $ {7^2}$, ...

Solution :

The list of numbers is $ {1^2} $, $ {3^2} $, $ {5^2} $, $ {7^2}$, .....

First term = $ { a_1 } $ = $ {1^2} $

Second term = $ { a_2 } $ = $ {3^2} $

Third term = $ { a_3 } = $ $ {5^2} $

Common difference d = $$ { a_2 - a_1 } = {3^2} - {1^2} $$

$$ = {9} - {1} $$

$$ = {8} $$

$$ { a_3 - a_2 } = {5^2} - {3^2} $$

$$ = {25} - {9} $$

$$ = 16 $$

We can clearly see that the difference between terms are not equal

$$ { a_3 - a_2 } \ne { a_2 - a_1 }$$

Thus given list of numbers do not form an Arithmetic Progression (AP)

Question 4 (xv)

Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
(xv) $ {1^2} $, $ {5^2} $, $ {7^2}$,73 ...

Solution :

The list of numbers is $ {1^2} $, $ {5^2} $, $ {7^2}$,73 .....

First term = $ { a_1 } $ = $ {1^2} $

Second term = $ { a_2 } $ = $ {5^2} $

Third term = $ { a_3 } = $ $ {7^2} $

Common difference d = $$ { a_2 - a_1 } = {5^2} - {1^2} $$

$$ = {25} - {1} $$

$$ = {24} $$

$$ { a_3 - a_2 } = {7^2} - {5^2} $$

$$ = {49} - {25} $$

$$ = 24 $$

$$ { a_4 - a_3 } = {73} - {7^2} $$

$$ = {73} - {49} $$

$$ = 24 $$

We can clearly see that the difference between terms are equal and equal to $ 24 $

$$ { a_3 - a_2 } = { a_2 - a_1 }$$

Thus given list of numbers form an Arithmetic Progression (AP)

We know that

$$ a_n = { a + ( n - 1) d } $$

∴ Next three terms :

$ { a_5 } = { {1^2} +(5-1) } × ( 24 ) = 97 $

$ { a_6 } = { {1^2} +(6-1) } × ( 24 ) = 121 $

$ { a_7 } = { {1^2} +(7-1) } × ( 24 ) = 145 $