Chapter 5: Arithmetic Progressions

Ans:  Given, the sum of the first $n$ terms of an A.P. is \[4n-{{n}^{2}}\].

First term $={{S}_{1}}=4-1=3$. …..(1)

Sum of first two terms $={{S}_{2}}=8-{{\left( 2 \right)}^{2}}=4$  …..(2) 

From (1) and (2), ${{2}^{nd}}$ term $={{S}_{2}}-{{S}_{1}}=4-3=1$. 

Sum of first three terms $={{S}_{3}}=12-{{\left( 3 \right)}^{2}}=3$  …..(3) 

From (3) and (2), ${{3}^{rd}}$ term $={{S}_{3}}-{{S}_{2}}=3-4=-1$. 

Similarly, 

Sum of first $n$ terms $={{S}_{n}}=4n-{{n}^{2}}$  …..(4) 

Sum of first $n-1$ terms $={{S}_{n-1}}=4\left( n-1 \right)-{{\left( n-1 \right)}^{2}}=-{{n}^{2}}+6n-5$  …..(5) 

From (4) and (5), ${{n}^{th}}$ term $={{S}_{n}}-{{S}_{n-1}}=\left( 4n-{{n}^{2}} \right)-\left( -{{n}^{2}}+6n-5 \right)=5-2n$  …..(6)

From (6), ${{10}^{th}}$ term is $5-2\left( 10 \right)=-15$. 

Ans:  First positive integer that is divisible by $6$ is $6$ itself. 

Second positive integer that is divisible by $6$ is \[6+6=12\].

Third positive integer that is divisible by $6$ is \[12+6=18\].  

Hence, it is an A.P. with first term and common difference both as $6$.  

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$     

Therefore, for $n=40$,

${{S}_{40}}=\dfrac{40}{2}\left[ 2\left( 6 \right)+6\left( 40-1 \right) \right]$

\[\Rightarrow {{S}_{40}}=120\left[ 41 \right]\] 

$\therefore {{S}_{40}}=4920$

Ans:  First positive integer that is divisible by $8$ is $8$ itself. 

Second positive integer that is divisible by $8$ is \[8+8=16\].

Third positive integer that is divisible by $8$ is \[16+8=24\].  

Hence, it is an A.P. with first term and common difference both as $8$.  

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$     

Therefore, for $n=15$,

${{S}_{15}}=\dfrac{15}{2}\left[ 2\left( 8 \right)+8\left( 15-1 \right) \right]$

\[\Rightarrow {{S}_{15}}=60\left[ 16 \right]\] 

$\therefore {{S}_{15}}=960$

Ans: The odd numbers between $0$ and $50$ are $1,3,5,...,49$. 

It is an A.P. with first term $1$ and common difference $2$.  ….(1)

We know that the ${{n}^{th}}$ term of the A.P. with first term $a$ and common difference $d$ is given by ${{a}_{n}}=a+\left( n-1 \right)d$  … (2)

Substitute ${{a}_{n}}=49$ and values from (1) into (2)

$49=1+2\left( n-1 \right)$

$\Rightarrow 24=\left( n-1 \right)$ 

$\therefore n=25$  ……(3)

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$  …..(4)

Substituting values from (1), (3) in (4) we get,

${{S}_{25}}=\dfrac{25}{2}\left[ 2+2\left( 25-1 \right) \right]$

\[\Rightarrow {{S}_{25}}=25\left[ 25 \right]\] 

$\therefore {{S}_{25}}=625$

Ans: Penalty of delay for first day is Rs. $200$.

Penalty of delay for second day is Rs. $250$.

Penalty of delay for third day is Rs. $300$.  

Hence it is an A.P. with first term $200$ and common difference $50$.

Money the contractor has to pay as penalty, if he has delayed the work by \[\mathbf{30}\] days is the sum of first $30$ terms of the A.P.

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. Therefore,

${{S}_{30}}=\dfrac{30}{2}\left[ 2\left( 200 \right)+50\left( 30-1 \right) \right]$

\[\Rightarrow {{S}_{30}}=15\left[ 400+50\left( 29 \right) \right]\] 

$\therefore {{S}_{30}}=27750$

Therefore, the contractor has to pay Rs \[27750\] as penalty.

Ans: Let the first prize be of Rs. $a$ then the second prize will be of Rs. $a-20$, the third prize will be of Rs. $a-40$.

Therefore, it is an A.P. with first term $a$ and common difference $-20$.

Given, ${{S}_{7}}=700$    

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. Therefore,

${{S}_{7}}=\dfrac{7}{2}\left[ 2a-20\left( 7-1 \right) \right]$

\[\Rightarrow 700=7\left[ a-60 \right]\] 

\[\Rightarrow 100=a-60\]

$\therefore a=160$

Therefore, the value of each of the prizes was \[Rs\text{ }160,\text{ }Rs\text{ }140,\text{ }Rs\text{ }120,Rs\text{ }100,\text{ }Rs\text{ }80,\text{ }Rs\text{ }60,\text{ }and\text{ }Rs\text{ }40.\] 

Ans: Each section of class I will plant $1$ tree each. Therefore, total trees planted by class I are $3$.  

Each section of class II will plant $2$ trees each. Therefore, total trees planted by class II are $3\times 2=6$.  

Each section of class III will plant $3$ trees each. Therefore, total trees planted by class III are $3\times 3=9$.  

Therefore, it is an A.P series with first term and common difference both as $3$. 

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. Therefore,

${{S}_{12}}=\dfrac{12}{2}\left[ 2\left( 3 \right)-3\left( 12-1 \right) \right]$

\[\Rightarrow {{S}_{12}}=6\left[ 39 \right]\] 

$\therefore {{S}_{12}}=234$

Therefore, $234$ trees will be planted by the students.

Ans: Length of first semi-circle ${{I}_{1}}=\pi \left( 0.5 \right)$ cm. 

Length of second semi-circle ${{I}_{2}}=\pi \left( 1 \right)$ cm.

Length of third semi-circle ${{I}_{3}}=\pi \left( 1.5 \right)$ cm.

Therefore, it is an A.P series with first term and common difference both as $\pi \left( 0.5 \right)$. 

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. Therefore,

${{S}_{13}}=\dfrac{13}{2}\left[ 2\left( 0.5\pi  \right)+\left( 0.5\pi  \right)\left( 13-1 \right) \right]$

\[\Rightarrow {{S}_{13}}=7\times 13\times \left( 0.5\pi  \right)\]

\[\Rightarrow {{S}_{13}}=7\times 13\times \dfrac{1}{2}\times \dfrac{22}{7}\] 

$\therefore {{S}_{13}}=143$

Therefore, the length of such spiral of thirteen consecutive semi-circles

will be \[143\] cm.

Ans: Total logs in first row are $20$.  

Total logs in second row are $19$.

Total logs in third row are $18$.

Therefore, it is an A.P series with first term $20$ and common difference $-1$. 

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. Therefore,

$200=\dfrac{n}{2}\left[ 2\left( 20 \right)-\left( n-1 \right) \right]$

\[\Rightarrow 400=n\left[ 41-n \right]\]

\[\Rightarrow {{n}^{2}}-41n+400=0\]

\[\Rightarrow {{n}^{2}}-16n-25n+400=0\]

\[\Rightarrow n\left( n-16 \right)-25\left( n-16 \right)=0\]

\[\Rightarrow \left( n-16 \right)\left( n-25 \right)=0\] 

For $n=25$, after ${{20}^{th}}$ term, all terms are negative, which is illogical as terms are representing the number of logs and number of logs being negative is illogical.

$\therefore n=16$ 

Total logs in ${{16}^{th}}$ row $=20-\left( 16-1 \right)=5$ 

Therefore, $200$ logs will be placed in $16$ rows and the total logs in ${{16}^{th}}$ row will be $5$.

Ans: Total distance run by competitor to collect and drop first potato $=2\times 5=10$m.

Total distance run by competitor to collect and drop second potato $=2\times \left( 5+3 \right)=16$m. 

Total distance run by competitor to collect and drop third potato $=2\times \left( 5+3+3 \right)=22$m. 

Therefore, it is an A.P series with first term $10$ and common difference $6$. 

We know that the sum of $n$ terms of the A.P. with first term $a$ and common difference $d$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. Therefore, to collect and drop $10$ potatoes total distance covered is

${{S}_{10}}=\dfrac{10}{2}\left[ 2\left( 10 \right)+6\left( 10-1 \right) \right]$

\[\Rightarrow {{S}_{10}}=5\left[ 74 \right]\]

$\therefore {{S}_{13}}=370$

Therefore, the competitor will run a total distance of \[370\]m.