Chapter 5: Arithmetic Progressions

Ans: Given the fare of first km is Rs.$15$ and the fare for each additional km is Rs. $8$. Hence, 

Taxi fare for ${{1}^{st}}$ km is Rs. $15$.  

Taxi fare for ${{2}^{nd}}$ km is Rs. $15+8=23$.  

Taxi fare for ${{3}^{rd}}$ km is Rs. \[23+8=31\].  

Similarly, Taxi fare for ${{n}^{th}}$ km is Rs. \[15+\left( n-1 \right)8\]. 

Therefore, we can conclude that the above list forms an A.P with common difference of $8$. 

1. The amount of air present in a cylinder when a vacuum pump removes a quarter of the air remaining in the cylinder at a time.

Ans: Let the initial volume of air in a cylinder be $V$ liter. In each stroke, the vacuum pump removes $\dfrac{1}{4}$ of air remaining in the cylinder at a time. Hence, 

Volume after ${{1}^{st}}$ stroke is $\dfrac{3V}{4}$.   

Volume after ${{2}^{nd}}$ stroke is $\dfrac{3}{4}\left( \dfrac{3V}{4} \right)$.   

Volume after ${{3}^{rd}}$ stroke is ${{\left( \dfrac{3}{4} \right)}^{2}}\left( \dfrac{3V}{4} \right)$.   

Similarly, Volume after ${{n}^{th}}$ stroke is ${{\left( \dfrac{3}{4} \right)}^{n}}V$.  

We can observe that the subsequent terms are not added with a constant digit but are being multiplied by $\dfrac{3}{4}$. Therefore, we can conclude that the above list does not forms an A.P. 

1. 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.

Ans: Given the cost of digging for the first meter is Rs.$150$ and the cost for each additional meter is Rs. $50$. Hence, 

Cost of digging for ${{1}^{st}}$ meter is Rs. $150$.  

Cost of digging for ${{2}^{nd}}$ meter is Rs. $150+50=200$.  

Cost of digging for ${{3}^{rd}}$ meter is Rs. \[200+50=250\].

Similarly, Cost of digging for ${{n}^{th}}$ meter is Rs. \[150+\left( n-1 \right)50\]. 

Therefore, we can conclude that the above list forms an A.P with common difference of $50$. 

1. The amount of money in the account every year, when Rs \[\mathbf{10000}\] is deposited at compound interest at \[\mathbf{8}\%\] per annum.

Ans: Given the principal amount is Rs.\[\mathbf{10000}\] and the compound interest is \[\mathbf{8}\%\] per annum. Hence,

Amount after ${{1}^{st}}$ year is Rs. $10000\left( 1+\dfrac{8}{100} \right)$.  

Amount after ${{2}^{nd}}$ year is Rs. $10000{{\left( 1+\dfrac{8}{100} \right)}^{2}}$.  

Amount after ${{3}^{rd}}$ year is Rs. $10000{{\left( 1+\dfrac{8}{100} \right)}^{3}}$.  

Similarly, Amount after ${{n}^{th}}$ year is Rs. $10000{{\left( 1+\dfrac{8}{100} \right)}^{n}}$. 

We can observe that the subsequent terms are not added with a constant digit but are being multiplied by \[\left( 1+\dfrac{8}{100} \right)\]. Therefore, we can conclude that the above list does not forms an A.P. 

Ans: We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$       …..(1)

Substituting \[a=10,d=10\] in (1) we get, ${{a}_{n}}=10+10\left( n-1 \right)=10n$ …..(2)

Therefore, from (2)

${{a}_{1}}=10$, ${{a}_{2}}=20$, ${{a}_{3}}=30$ and ${{a}_{4}}=40$.

1. \[a=-2,d=0\]

Ans: We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$       …..(1)

Substituting \[a=-2,d=0\] in (1) we get, ${{a}_{n}}=-2+0\left( n-1 \right)=-2$ …..(2)

Therefore, from (2)

${{a}_{1}}=-2$, ${{a}_{2}}=-2$, ${{a}_{3}}=-2$ and ${{a}_{4}}=-2$.

1. \[a=4,d=-3\] 

Ans: We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$       …..(1)

Substituting \[a=4,d=-3\] in (1) we get, ${{a}_{n}}=4-3\left( n-1 \right)=7-3n$ …..(2)

Therefore, from (2)

${{a}_{1}}=4$, ${{a}_{2}}=1$, ${{a}_{3}}=-2$ and ${{a}_{4}}=-5$.

1. \[a=-1\text{,}d=1/2\] 

Ans: We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$       …..(1)

Substituting \[a=-1\text{,}d=1/2\] in (1) we get, ${{a}_{n}}=-1+\dfrac{1}{2}\left( n-1 \right)=\dfrac{n-3}{2}$ …..(2)

Therefore, from (2)

${{a}_{1}}=-1$, ${{a}_{2}}=-\dfrac{1}{2}$, ${{a}_{3}}=0$ and ${{a}_{4}}=\dfrac{1}{2}$.

1. \[a=-1.25,d=-0.25\] 

Ans: We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$       …..(1)

Substituting \[a=-1.25,d=-0.25\] in (1) we get, ${{a}_{n}}=-1.25-0.25\left( n-1 \right)=-1-0.25n$ …..(2)

Therefore, from (2)

${{a}_{1}}=-1.25$, ${{a}_{2}}=-1.5$, ${{a}_{3}}=-1.75$ and ${{a}_{4}}=-2$.

Ans: From the given AP, we can see that the first term is $3$.

The common difference is the difference between any two consecutive numbers of the A.P. 

Common difference $=$ ${{2}^{nd}}\text{ }term-{{1}^{st}}\ term$

$\therefore $ Common difference $=$ $1-3=-2$.  

1. \[-5,-1,3,7,...\] 

Ans: From the given AP, we can see that the first term is $-5$.

The common difference is the difference between any two consecutive numbers of the A.P. 

Common difference $=$ ${{2}^{nd}}\text{ }term-{{1}^{st}}\ term$

$\therefore $ Common difference $=$ $-1-\left( -5 \right)=4$.  

1. \[\dfrac{1}{3}\text{,}\dfrac{5}{3}\text{,}\dfrac{9}{3}\text{,}\dfrac{13}{3},...\]

Ans: From the given AP, we can see that the first term is $\dfrac{1}{3}$.

The common difference is the difference between any two consecutive numbers of the A.P. 

Common difference $=$ ${{2}^{nd}}\text{ }term-{{1}^{st}}\ term$

$\therefore $ Common difference $=$ $\dfrac{5}{3}-\dfrac{1}{3}=\dfrac{4}{3}$.  

1. \[0.6,1.7,2.8,3.9,...\] 

Ans: From the given AP, we can see that the first term is $0.6$.

The common difference is the difference between any two consecutive numbers of the A.P. 

Common difference $=$ ${{2}^{nd}}\text{ }term-{{1}^{st}}\ term$

$\therefore $ Common difference $=$ $1.7-0.6=1.1$.  

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

${{a}_{2}}-{{a}_{1}}=4-2=2$              …..(1)

${{a}_{3}}-{{a}_{2}}=8-4=4$              …..(2)

${{a}_{4}}-{{a}_{3}}=16-8=8$              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.

Therefore, the given series does not form an A.P.

1. \[\text{2,}\dfrac{5}{2}\text{,3,}\dfrac{7}{2}\text{,}...\] 

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

${{a}_{2}}-{{a}_{1}}=\dfrac{5}{2}-2=\dfrac{1}{2}$              …..(1)

${{a}_{3}}-{{a}_{2}}=3-\dfrac{5}{2}=\dfrac{1}{2}$              …..(2)

${{a}_{4}}-{{a}_{3}}=\dfrac{7}{2}-3=\dfrac{1}{2}$              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term $2$ and common difference $\dfrac{1}{2}$. 

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$       …..(4)

Substituting \[a=2,d=\dfrac{1}{2}\] in (4) we get, ${{a}_{n}}=2+\dfrac{1}{2}\left( n-1 \right)=\dfrac{n+3}{2}$ …..(5)

Therefore, from (5)

${{a}_{5}}=4$, ${{a}_{6}}=\dfrac{9}{2}$ and ${{a}_{7}}=5$.

1. \[1.2,3.2,5.2,7.2...\] 

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

${{a}_{2}}-{{a}_{1}}=3.2-1.2=2$              …..(1)

${{a}_{3}}-{{a}_{2}}=5.2-3.2=2$              …..(2)

${{a}_{4}}-{{a}_{3}}=7.2-5.2=2$              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term $1.2$ and common difference $2$. 

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$       …..(4)

Substituting \[a=1.2,d=2\] in (4) we get, ${{a}_{n}}=1.2+2\left( n-1 \right)=2n-0.8$ …..(5)

Therefore, from (5)

${{a}_{5}}=9.2$, ${{a}_{6}}=11.2$ and \[{{a}_{7}}=13.2\].

1. $-10,-6,-2,2,...$ 

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

${{a}_{2}}-{{a}_{1}}=-6-\left( -10 \right)=4$              …..(1)

${{a}_{3}}-{{a}_{2}}=-2-\left( -6 \right)=4$              …..(2)

${{a}_{4}}-{{a}_{3}}=2-\left( -2 \right)=4$              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term $-10$ and common difference $4$. 

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$       …..(4)

Substituting \[a=-10,d=4\] in (4) we get, ${{a}_{n}}=-10+4\left( n-1 \right)=4n-14$ …..(5)

Therefore, from (5)

${{a}_{5}}=6$, ${{a}_{6}}=10$ and \[{{a}_{7}}=14\].

1. \[3,3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2},...\] 

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

\[{{a}_{2}}-{{a}_{1}}=\left( 3+\sqrt{2} \right)-\left( 3 \right)=\sqrt{2}\]              …..(1)

${{a}_{3}}-{{a}_{2}}=\left( 3+2\sqrt{2} \right)-\left( 3+\sqrt{2} \right)=\sqrt{2}$              …..(2)

${{a}_{4}}-{{a}_{3}}=\left( 3+3\sqrt{2} \right)-\left( 3+2\sqrt{2} \right)=\sqrt{2}$              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term $3$ and common difference \[\sqrt{2}\]. 

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$       …..(4)

Substituting \[a=3,d=\sqrt{2}\] in (4) we get, ${{a}_{n}}=3+\left( n-1 \right)\sqrt{2}$   …..(5)

Therefore, from (5)

${{a}_{5}}=3+4\sqrt{2}$, ${{a}_{6}}=3+5\sqrt{2}$ and \[{{a}_{7}}=3+6\sqrt{2}\].

1. \[\text{0}\text{.2,0}\text{.22,0}\text{.222,0}\text{.2222}.....\] 

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

${{a}_{2}}-{{a}_{1}}=0.22-0.2=0.02$              …..(1)

${{a}_{3}}-{{a}_{2}}=0.222-0.22=0.002$              …..(2)

${{a}_{4}}-{{a}_{3}}=0.2222-0.222=0.0002$              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.

Therefore, the given series does not form an A.P.

1. \[0,-4,-8,-12....\] 

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

${{a}_{2}}-{{a}_{1}}=-4-0=-4$              …..(1)

${{a}_{3}}-{{a}_{2}}=-8-\left( -4 \right)=-4$              …..(2)

${{a}_{4}}-{{a}_{3}}=-12-\left( -8 \right)=-4$              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term $0$ and common difference $-4$. 

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$       …..(4)

Substituting \[a=0,d=-4\] in (4) we get, ${{a}_{n}}=0-4\left( n-1 \right)=4-4n$ …..(5)

Therefore, from (5)

${{a}_{5}}=-16$, ${{a}_{6}}=-20$ and \[{{a}_{7}}=-24\].

1. $-\dfrac{1}{2},-\dfrac{1}{2},-\dfrac{1}{2},-\dfrac{1}{2}....$

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

${{a}_{2}}-{{a}_{1}}=\left( -\dfrac{1}{2} \right)-\left( -\dfrac{1}{2} \right)=0$              …..(1)

${{a}_{3}}-{{a}_{2}}=\left( -\dfrac{1}{2} \right)-\left( -\dfrac{1}{2} \right)=0$              …..(2)

${{a}_{4}}-{{a}_{3}}=\left( -\dfrac{1}{2} \right)-\left( -\dfrac{1}{2} \right)=0$              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term $-\dfrac{1}{2}$ and common difference $0$. 

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$       …..(4)

Substituting \[a=-\dfrac{1}{2},d=0\] in (4) we get, ${{a}_{n}}=-\dfrac{1}{2}+0\left( n-1 \right)=-\dfrac{1}{2}$ …..(5)

Therefore, from (5)

${{a}_{5}}=-\dfrac{1}{2}$, ${{a}_{6}}=-\dfrac{1}{2}$ and \[{{a}_{7}}=-\dfrac{1}{2}\].

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

${a_2} - {a_1} = 3 - 1 = 2$         …..(1)

${a_3} - {a_2} = 9 - 3 = 6$           …..(2)

${a_4} - {a_3} = 27 - 9 = 18$           …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

${a_2} - {a_1} = 2a - a = a$              …..(1)

${a_3} - {a_2} = 3a - 2a = a$              …..(2)

${a_4} - {a_3} = 4a - 3a = a$              …..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term $a$ and common difference $a$. 

We know that the ${n^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${a_n} = a + \left( {n - 1} \right)d$       …..(4)

Substituting, $a = a,d = a$in (4)

we get, ${a_n} = a + \left( {n - 1} \right)d$....(5)

${a_5} = a + (5 - 1)a = 5a$

${a_6} = a + (6 - 1)a = 6a$

${a_7} = a + (7 - 1)a = 7a$

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.${a_2} - {a_1} = {a^2} - a = a\left( {a - 1} \right)$              …..(1)

${a_3} - {a_2} = {a^3} - {a^2} = {a^2}\left( {a - 1} \right)$         …..(2)

${a_4} - {a_3} = {a^4} - {a^3} = {a^3}\left( {a - 1} \right)$  ……..(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.

Therefore, the given series does not form an A.P.

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

${a_2} - {a_1} = \sqrt 8  - \sqrt 2  = 2\sqrt 2  - \sqrt 2  = \sqrt 2 $     ……(1)

${a_3} - {a_2} = \sqrt {18}  - \sqrt 8  = 3\sqrt 2  - 2\sqrt 2  = \sqrt 2 $   ………(2)

${a_4} - {a_3} = \sqrt {32}  - \sqrt {18}  = 4\sqrt 2  - 3\sqrt 2  = \sqrt 2 $.......(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term $a$ and common difference is given by${a_n} = a + (n - 1)d$      …..(4)

Substituting, a = $\sqrt 2 $d = $\sqrt 2 $ in (4)

we get, 

${a_5} = \sqrt 2  + \left( {5 - 1} \right)\sqrt 2  = \sqrt 2  + 4\sqrt 2  = 5\sqrt 2  = \sqrt {50} $

${a_6} = \sqrt 2  + \left( {6 - 1} \right)\sqrt 2  = \sqrt 2  + 5\sqrt 2  = 6\sqrt 2  = \sqrt {72} $

${a_7} = \sqrt 2  + \left( {7 - 1} \right)\sqrt 2  = \sqrt 2  + 6\sqrt 2  = 7\sqrt 2  = \sqrt {98} $

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

${a_2} - {a_1} = \sqrt 6  - \sqrt 3  = \sqrt 3  \times \sqrt 2  - \sqrt 3  = \sqrt 3 \left( {\sqrt 2  - 1} \right)$     ……(1)

${a_3} - {a_2} = \sqrt 9  - \sqrt 6  = 3 - \sqrt 6  = \sqrt 3 \left( {\sqrt 3  - \sqrt 2 } \right)$   ………(2)

${a_4} - {a_3} = \sqrt {12}  - \sqrt 9  = 2\sqrt 3  - 3 = \sqrt 3 \left( {2 - \sqrt 3 } \right)$……….(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.

Therefore, the given series does not form an A.P.

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

${a_2} - {a_1} = {3^2} - {1^2} = 8$     ……(1)

${a_3} - {a_2} = {5^2} - {3^2} = 16$  ………(2)

${a_4} - {a_3} = {7^2} - {5^2} = 24$……….(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is not equal.

Therefore, the given series does not form an A.P.

Ans: For the given series, let us check the difference between all consecutive terms and find if they are equal or not.

${a_2} - {a_1} = {5^2} - {1^2} = 24$ ……(1)

${a_3} - {a_2} = {7^2} - {5^2} = 24$………(2)

${a_4} - {a_3} = 73 - {7^2} = 24$  ……….(3)

From (1), (2), and (3) we can see that the difference between all consecutive terms is equal.

Therefore, the given series form an A.P. with first term a and common difference is given by${a_n} = a + (n - 1)d$      …..(4)

Substituting a = ${1^2}$, d=24 in (4)

we get, 

${a_5} = {1^2} + \left( {5 - 1} \right)24 = 1 + \left( 4 \right)24 = 97$

${a_6} = {1^2} + \left( {6 - 1} \right)24 = 1 + \left( 5 \right)24 = 121$

${a_7} = {1^2} + \left( {7 - 1} \right)24 = 1 + \left( 6 \right)24 = 145$