Exercise 5.4 (Revised) - Chapter 5 - Arithmetic Equations - Ncert Solutions class 10 - Maths

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NCERT Solutions Class 10 Maths Chapter 5: Arithmetic Equations

Ex 5.4 Question 1.

Which term of the AP: $121,117,113, \ldots .$. is its first negative term?

Answer.

Given: $121,117,113, \ldots \ldots$
Here $a=121, d=117-121=-4$
$
\begin{aligned}
& \text { Now, } a_n=a+(n-1) d \\
& =121+(n-1)(-4) \\
& =121-4 n+4=125-4 n
\end{aligned}
$

For the first negative term, $a_n<0$
$
\begin{aligned}
& \Rightarrow 125-4 n<0 \\
& \Rightarrow 125<4 n \\
& \Rightarrow \frac{125}{4} & \Rightarrow 31 \frac{1}{4} & n \text { is an integer and } n>31 \frac{1}{4}
\end{aligned}
$
$n$ is an integer and $n>31 \frac{1}{4}$.
Hence, the first negative term is $32^{\text {nd }}$ term.
Ex 5.4 Question 2.

The sum of the third and the seventh terms of an AP is 6 and their product is 8 . Find the sum of sixteen terms of the AP.

Answer.

Let the AP be $a-4 d, a-3 d, a-2 d, a-d, a, a+d, a+2 d, a+3 d, \ldots$.
Then, $a_3=a-2 d, a_7=a+2 d$
$
\begin{aligned}
& \Rightarrow a_3+a_7=a-2 d+a+2 d=6 \\
& \Rightarrow 2 a=6 \\
& \Rightarrow a=3 \ldots \ldots \ldots . \text { (i) }
\end{aligned}
$

Also $(a-2 d)(a+2 d)=8$
$
\begin{aligned}
& \Rightarrow a^2-4 d^2=8 \\
& \Rightarrow 4 d^2=a^2-8 \\
& \Rightarrow 4 d^2=3^2-8 \\
& \Rightarrow 4 d^2=1 \\
& \Rightarrow d^2=\frac{1}{4} \Rightarrow d= \pm \frac{1}{2}
\end{aligned}
$

Taking $d=\frac{1}{2}$,
$
\begin{aligned}
& S_{16}=\frac{16}{2}[2 \times(a-4 d)+(16-1) d] \\
& =8\left[2 \times\left(3-4 \times \frac{1}{2}\right)+15 \times \frac{1}{2}\right] \\
& =8\left[2+\frac{15}{2}\right]=8 \times \frac{19}{2}=76
\end{aligned}
$

$\text { Taking } d=\frac{-1}{2} \text {, }$

$
\begin{aligned}
& S_{16}=\frac{16}{2}[2 \times(a-4 d)+(16-1) d] \\
& =8\left[2 \times\left(3-4 \times \frac{-1}{2}\right)+15 \times \frac{-1}{2}\right] \\
& =8\left[\frac{20-15}{2}\right]=8 \times \frac{5}{2}=20 \\
& \therefore S_{16}=20 \text { and } 76
\end{aligned}
$
Ex 5.4 Question 3.

A ladder has rungs $25 \mathrm{~cm}$ apart (see figure). The rungs decrease uniformly in length from $45 \mathrm{~cm}$, at the bottom to $25 \mathrm{~cm}$ at the top. If the top and the bottom rungs are $2 \frac{1}{2}$ $\mathrm{m}$ apart, what is the length of the wood required for the rungs?

Answer.

Number of rungs $(n)=\frac{2 \frac{1}{2} \times 100}{25}=10$
The length of the wood required for rungs = sum of 10 rungs
$
=\frac{10}{2}[25+45]=5 \times 70=350 \mathrm{~cm}
$

Ex 5.4 Question 4.

The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of $x$ such that the sum of the numbers of the houses preceding the house numbered $x$ is equal to the sum of the numbers of the houses following it. Find this value of $x$.

Answer.

Here $a=1$ and $d=1$
$
\begin{aligned}
& \therefore S_{x-1}=\frac{x-1}{2}[2 \times 1+(x-1-1) \times 1] \\
& =\frac{x-1}{2}(2+x-2) \\
& \frac{(x-1) x}{2}=\frac{x^2-x}{2} \\
& S_x=\frac{x}{2}[2 \times 1+(x-1) \times 1] \\
& =\frac{x}{2}(x+1)=\frac{x^2+x}{2} \\
& S_{49}=\frac{49}{2}[2 \times 1+(49-1) \times 1] \\
& =\frac{49}{2}(2+48)=49 \times 25
\end{aligned}
$

According to question,
$
\begin{aligned}
& S_{x-1}=S_{49}-S_x \\
& \Rightarrow \frac{x^2-x}{2}=49 \times 25-\frac{x^2+x}{2} \\
& \Rightarrow \frac{x^2-x}{2}+\frac{x^2+x}{2}=49 \times 25 \\
& \Rightarrow \frac{x^2-x+x^2+x}{2}=49 \times 25 \\
& \Rightarrow x^2=49 \times 25
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow x^2=49 \times 25 \\
& \Rightarrow x= \pm 35
\end{aligned}
$

Since, $x$ is a counting number, so negative value will be neglected.
$
\therefore x=35
$

Ex 5.4 Question 5.

A small terrace at a football ground comprises of 15 steps each of which is $50 \mathrm{~m}$ long and built of solid concrete.

Each step has a rise of $\frac{1}{4} \mathrm{~m}$ and a tread of $\frac{1}{2} \mathrm{~m}$ (see figure). Calculate the total volume of concrete required to build the terrace.

Answer.

Volume of concrete required to build the first step, second step, third step, ....... (in $\mathrm{m}^2$ ) are
$
\begin{aligned}
& \frac{1}{4} \times \frac{1}{2} \times 50,\left(2 \times \frac{1}{4}\right) \times \frac{1}{2} \times 50,\left(3 \times \frac{1}{4}\right) \times \frac{1}{2} \times 50, \ldots . \\
& \Rightarrow \frac{50}{8}, 2 \times \frac{50}{8}, 3 \times \frac{50}{8}, \ldots \ldots . \\
& \therefore \text { Total volume of concrete required }=\frac{50}{8}+2 \times \frac{50}{8}+3 \times \frac{50}{8}+\ldots \ldots . \\
& =\frac{50}{8}[1+2+3+\ldots \ldots .] \\
& =\frac{50}{8} \times \frac{15}{2}[2 \times 1+(15-1) \times 1][\because n=15] \\
& =\frac{50}{8} \times \frac{15}{2} \times 16 \\
& =750 \mathrm{~m}^3
\end{aligned}
$