NCERT Solutions for Exercise 5.4 Class 10 Maths Chapter 5 - Arithmetic Progressions

Answer:

Given AP is
$121,117,113,...,$
Here $a=121andd=-4$
Let suppose nth term of the AP is first negative term
Then,
$a_n=a+(n-1)d$
If nth term is negative then $a_n<0$
$\Rightarrow 121+(n-1)(-4)<0$
$\Rightarrow 125<4n$
$\Rightarrow n>\frac{125}{4}=31.25$
Therefore, first negative term must be 32nd term

Q2 The sum of the third and the seventh terms of an AP is $6$ and their product is $8$ . Find the sum of first sixteen terms of the AP.

Answer:

It is given that sum of third and seventh terms of an AP are and their product is $8$
$a_3=a+2d$
$a_7=a+6d$
Now,
$a_3+a_7=a+2d+a+6d=6$
$\Rightarrow 2a+8d=6$
$\Rightarrow a+4d=3\Rightarrow a=3-4d-(i)$
And
$a_3.a_7=(a+2d).(a+6d)=a^2+8ad+12d^2=8-(ii)$
put value from equation (i) in (ii) we will get
$\Rightarrow (3-4d{)}^2+8(3-4d)d+12d^2=8$
$\Rightarrow 9+16d^2-24d+24d-32d^2+12d^2=8$
$\Rightarrow 4d^2=1$
$\Rightarrow d=\pm \frac{1}{2}$
Now,
case (i) $d=\frac{1}{2}$

$a=3-4\times \frac{1}{2}=1$
Then,
$S_{16}=\frac{16}{2}\{2\times 1+(16-1)\frac{1}{2}\}$

$S_{16}=76$

case (ii) $d=-\frac{1}{2}$
$a=3-4\times (-\frac{1}{2})=5$
Then,
$S_{16}=\frac{16}{2}\{2\times 1+(16-1)(-\frac{1}{2})\}$

$S_{16}=20$

Q3 A ladder has rungs $25$ cm apart. (see Fig. $5.7$ ). The rungs decrease uniformly in length from $45$ cm at the bottom to $25$ cm at the top. If the top and the bottom rungs are $2\frac{1}{2}$ m apart, what is the length of the wood required for the rungs? [ Hint: Number of rungs $=\frac{250}{25}+1]$

Answer:

It is given that
The total distance between the top and bottom rung $=2\frac{1}{2}m=250cm$
Distance between any two rungs = 25 cm
Total number of rungs = $\frac{250}{25}+1=11$
And it is also given that bottom-most rungs is of 45 cm length and topmost is of 25 cm length.As it is given that the length of rungs decrease uniformly, it will form an AP with $a=25,a_{11}=45andn=11$
Now, we know that
$a_{11}=a+10d$

$\Rightarrow 45=25+10d$
$\Rightarrow d=2$
Now, total length of the wood required for the rungs is equal to
$S_{11}=\frac{11}{2}\{2\times 25+(11-1)2\}$
$S_{11}=\frac{11}{2}\{50+20\}$
$S_{11}=\frac{11}{2}\times 70$
$S_{11}=385cm$
Therefore, the total length of the wood required for the rungs is equal to 385 cm

Q4 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 $x$ . [ Hint : $\text{Double subscripts: use braces to clarify}$ ]

Answer:

It is given 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
And 1,2,3,.....,49 form an AP with a = 1 and d = 1
Now, we know that
$S_n=\frac{n}{2}\{2a+(n-1)d\}$
Suppose their exist an n term such that ( n < 49)
Now, according to given conditions
Sum of first n - 1 terms of AP = Sum of terms following the nth term
Sum of first n - 1 term of AP = Sum of whole AP - Sum of first m terms of AP
i.e.
$S_{n-1}=S_{49}-S_n$

$\frac{n-1}{2}\{2a+((n-1)-1)d\}=\frac{49}{2}\{2a+(49-1)d\}-\frac{n}{2}\{2a+(n-1)d\}$

$\frac{n-1}{2}\{2+(n-2)\}=\frac{49}{2}\{2+48\}-\frac{n}{2}\{2+(n-1)\}$

$\frac{n-1}{2}\left\{n\right\}=\frac{49}{2}\left\{50\right\}-\frac{n}{2}\{n+1\}$

$\frac{n^2}{2}-\frac{n}{2}=1225-\frac{n^2}{2}-\frac{n}{2}$

$n^2=1225$
$n=\pm 35$

Given House number are not negative so we reject n = -35

Therefore, the sum of no of houses preceding the house no 35 is equal to the sum of no of houses following the house no 35

Q5 A small terrace at 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 $\frac{1}{4}m$ and a tread of $\frac{1}{2}m$ . (see Fig. $5.8$ ). Calculate the total volume of concrete required to build the terrace.
[ Hint: Volume of concrete required to build the first step $=\frac{1}{4}\times \frac{1}{2}\times 50m^3$ ]

Answer:

It is given that
football ground comprises of $15$ steps each of which is $50$ m long and Each step has a rise of $\frac{1}{4}m$ and a tread of $\frac{1}{2}m$
Now,
The volume required to make the first step = $\frac{1}{4}\times \frac{1}{2}\times 50=6.25m^3$

Similarly,

The volume required to make 2nd step = $(\frac{1}{4}+\frac{1}{4})\times \frac{1}{2}\times 50=\frac{1}{2}\times \frac{1}{2}\times 50=12.5m^3$
And
The volume required to make 3rd step = $(\frac{1}{4}+\frac{1}{4}+\frac{1}{4})\times \frac{1}{2}\times 50=\frac{3}{4}\times \frac{1}{2}\times 50=18.75m^3$

And so on
We can clearly see that this is an AP with $a=6.25andd=6.25$
Now, the total volume of concrete required to build the terrace of 15 such step is
$S_{15}=\frac{15}{2}\{2\times 6.25+(15-1)6.25\}$
$S_{15}=\frac{15}{2}\{12.5+87.5\}$
$S_{15}=\frac{15}{2}\times 100$
$S_{15}=15\times 50=750$
Therefore, the total volume of concrete required to build the terrace of 15 such steps is $750m^3$