Chapter 5: Arithmetic Progressions

Ans: Given A.P. \[\mathbf{121},\mathbf{117},\mathbf{113},...\] 

Its first term is $121$ and the common difference is $117-121=-4$. 

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

Therefore the ${{n}^{th}}$ term of the given A.P. is ${{a}_{n}}=121-4\left( n-1 \right)$   ….. (1)

To find negative term, find $n$ such that ${{a}_{n}} <0$  

Hence from (1),

$121-4\left( n-1 \right) <0$

$\Rightarrow 121 <4\left( n-1 \right)$ 

\[\Rightarrow \dfrac{121}{4}+1 < n\]

$\Rightarrow n >\dfrac{125}{4}$ 

$\therefore n >31.25$ 

Therefore, the ${{32}^{nd}}$ term of the given A.P. will be its first negative term.

Ans: Given the sum of third and seventh term of A.P., ${{a}_{3}}+{{a}_{7}}=6$  …..(1)

Given the sum of third and seventh term of A.P., ${{a}_{3}}\cdot {{a}_{7}}=8$  …..(2)

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

For $n=3,{{a}_{3}}=a+2d$ 

For $n=7,{{a}_{7}}=a+6d$ 

From (1), ${{a}_{3}}+{{a}_{7}}=\left( a+2d \right)+\left( a+6d \right)$ 

$\Rightarrow 2a+8d=6$ 

$\therefore a+4d=3$   ….. (3)

From (2), ${{a}_{3}}\cdot {{a}_{7}}=\left( a+2d \right)\cdot \left( a+6d \right)$ 

$\therefore {{a}^{2}}+8ad+12{{d}^{2}}=8$   …..(4)

Let us now solve equations (3) and (4) by substituting the value of $a$ from (3) into (4).

${{\left( 3-4d \right)}^{2}}+8d\left( 3-4d \right)+12{{d}^{2}}=8$

$\Rightarrow 9-24d+16{{d}^{2}}+24d-32{{d}^{2}}+12{{d}^{2}}=8$ 

$\Rightarrow -4{{d}^{2}}+1=0$

$\Rightarrow {{d}^{2}}=\dfrac{1}{4}$

$\therefore d=\dfrac{1}{2},-\dfrac{1}{2}$ …..(5)

CASE 1: 

For $d=\dfrac{1}{2}$ 

 Substitute $d=\dfrac{1}{2}$ in (6) we get, $a=1$ …..(6)

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

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, 

${{S}_{16}}=\dfrac{16}{2}\left[ 2+\dfrac{1}{2}\left( 16-1 \right) \right]$

$\Rightarrow {{S}_{16}}=4\left[ 19 \right]$ 

$\therefore {{S}_{16}}=76$ 

CASE 2: 

For $d=-\dfrac{1}{2}$ 

 Substitute $d=-\dfrac{1}{2}$ in (6) we get, $a=5$ …..(7)

Therefore, it is an A.P series with first term $5$ and common difference $-\dfrac{1}{2}$ and hence, 

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

$\Rightarrow {{S}_{16}}=\dfrac{16}{2}\left[ 2\left( 5 \right)-\dfrac{1}{2}\left( 16-1 \right) \right]$

$\Rightarrow {{S}_{16}}=4\left[ 5 \right]$ 

$\therefore {{S}_{16}}=20$ 

Ans: Distance between first and last rungs is $2\dfrac{1}{2}m=\dfrac{5}{2}m=250cm$.

Distance between two consecutive rungs is $25cm$.

Therefore, total number of rungs are $\dfrac{250}{25}+1=11$.

Also, we can observe that the length of each rung is decreasing in a uniform order. So, we can conclude that the length of rungs is in A.P. with first term $45$, common difference $-25$ and number of terms $11$.   

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$. Therefore, 

${{S}_{11}}=\dfrac{11}{2}\left[ 45+25 \right]$

$\Rightarrow {{S}_{11}}=11\left[ 35 \right]$ 

$\therefore {{S}_{11}}=385$

Therefore, the length of the wood required for the rungs is \[385\]cm.

Ans: Given houses are numbered $1,2,3,4,....$

Clearly, they are numbered in A.P. series with both first term and common difference as $1$.

Now, there is house numbered $x$ such that the sum of numbers of the houses preceding the house numbered \[x\] is equal to the sum of the number of houses following it i.e., \[{{S}_{x-1}}={{S}_{49}}-{{S}_{x}}\]

We know that the sum of $n$ terms of the A.P. with first term $a$ and last term $l$ is given by ${{S}_{n}}=\dfrac{n}{2}\left[ a+l \right]$. Therefore, 

\[{{S}_{x-1}}={{S}_{49}}-{{S}_{x}}\]

$\Rightarrow \left\{ \dfrac{\left( x-1 \right)}{2}\left[ 1+\left( x-1 \right) \right] \right\}=\left\{ \dfrac{49}{2}\left[ 1+49 \right] \right\}-\left\{ \dfrac{x}{2}\left[ 1+x \right] \right\}$

$\Rightarrow \dfrac{x\left( x-1 \right)}{2}=49\left[ 25 \right]-\dfrac{x\left( x+1 \right)}{2}$ 

$\Rightarrow x\left( x-1 \right)=2450-x\left( x+1 \right)$

$\Rightarrow 2{{x}^{2}}=2450$

$\therefore x=35$   (Since house number cannot be negative)

Therefore, house number \[35\] is such that the sum of the numbers of houses preceding the house numbered \[35\] is equal to the sum of the numbers of the houses following it.

Ans: Given that a football ground comprises of \[15\] steps each of which is \[50\] m long and built of solid concrete. Each step has a rise of $\dfrac{1}{4}$m and a tread of $\dfrac{1}{2}$m. An easy illustration of the problem is depicted below.

Here blue step is the lowermost step. Let it be known as step $1$. The volume of step $1$ is $\dfrac{1}{2}\times \dfrac{1}{4}\times 50\text{ }{{m}^{3}}$.

The red step is the second lowermost step. Let it be known as step $2$. The volume of step $2$ is $\dfrac{1}{2}\times \dfrac{1}{2}\times 50\text{ }{{m}^{3}}$.

The green step is the third lower step. Let it be known as step $3$. The volume of step $3$ is $\dfrac{1}{2}\times 1\times 50\text{ }{{m}^{3}}$.  

We can see that the height is increasing with each increasing step by a factor of $\dfrac{1}{4}$, length and width being constant. Hence the volume of each step is increasing by $\dfrac{1}{2}\times \dfrac{1}{4}\times 50\text{ }{{m}^{3}}$.

Therefore, we can conclude that the volume of steps is in A.P. with first term and common difference both as $\dfrac{1}{2}\times \dfrac{1}{4}\times 50=\dfrac{25}{4}\text{ }{{m}^{3}}$.

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, 

${{S}_{15}}=\dfrac{15}{2}\left[ 2\left( \dfrac{25}{4} \right)+\left( \dfrac{25}{4} \right)\left( 15-1 \right) \right]$

$\Rightarrow {{S}_{15}}=\dfrac{15}{2}\left( \dfrac{25}{4} \right)\left[ 16 \right]$ 

$\Rightarrow {{S}_{15}}=15\cdot 25\cdot 2$

$\therefore {{S}_{15}}=750$

Therefore, volume of concrete required to build the terrace is $750\text{ }{{m}^{3}}$.