EXERCISE 5.4 (OPTIONAL)

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Question 1

Question:

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

Solution:

Given,

$$ a=121,\qquad d=117-121=-4 $$

The nth term of an AP is

$$ a_n=a+(n-1)d $$

Substituting the given values,

$$ a_n=121+(n-1)(-4) $$

$$ a_n=121-4n+4 $$

$$ a_n=125-4n $$

For the first negative term,

$$ 125-4n<0 $$

$$ 125<4n $$

$$ n>31.25 $$

The smallest natural number satisfying this condition is

$$ n=32 $$

Verification:

$$ a_{31}=125-4(31)=1 $$

$$ a_{32}=125-4(32)=-3 $$

Hence, the first negative term is the 32nd term.

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Answer

$$ \boxed{\text{The first negative term is the }32^{\text{nd}}\text{ term}.} $$

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Question 2

Question:

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

Solution:

Let the first term be $a$ and the common difference be $d$.

Third term:

$$ a_3=a+2d $$

Seventh term:

$$ a_7=a+6d $$

Given,

$$ (a+2d)+(a+6d)=6 $$

$$ 2a+8d=6 $$

$$ a+4d=3 $$

Also,

$$ (a+2d)(a+6d)=8 $$

Since

$$ a+2d=(a+4d)-2d=3-2d $$

and

$$ a+6d=(a+4d)+2d=3+2d $$

Therefore,

$$ (3-2d)(3+2d)=8 $$

Using the identity

$$ (x-y)(x+y)=x^2-y^2, $$

we get

$$ 9-4d^2=8 $$

$$ 4d^2=1 $$

$$ d=\pm\frac12 $$

Now,

If $$ d=\frac12, $$ then

$$ a=3-4\left(\frac12\right)=1 $$

If $$ d=-\frac12, $$ then

$$ a=3-4\left(-\frac12\right)=5 $$

The sum of the first 16 terms is

$$ S_{16}=\frac{16}{2}\left[2a+15d\right] $$

Case 1:

$$ a=1,\qquad d=\frac12 $$

$$ S_{16}=8\left[2(1)+15\left(\frac12\right)\right] $$

$$ =8\left[2+\frac{15}{2}\right] $$

$$ =8\left(\frac{19}{2}\right) $$

$$ =76 $$

Case 2:

$$ a=5,\qquad d=-\frac12 $$

$$ S_{16}=8\left[2(5)+15\left(-\frac12\right)\right] $$

$$ =8\left[10-\frac{15}{2}\right] $$

$$ =8\left(\frac52\right) $$

$$ =20 $$

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Answer

The AP can be:

$$ a=1,\qquad d=\frac12 $$

Hence,

$$ \boxed{S_{16}=76} $$

or

$$ a=5,\qquad d=-\frac12 $$

Hence,

$$ \boxed{S_{16}=20} $$

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Question 3

Question:

A ladder has rungs 25 cm apart. 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 $$ 2\frac{1}{2}\text{ m} $$ apart, what is the length of the wood required for the rungs?

Solution:

Distance between the top and bottom rungs is

$$ 2\frac{1}{2}\text{ m}=250\text{ cm} $$

Distance between two consecutive rungs is

$$ 25\text{ cm} $$

Therefore, the number of gaps between the rungs is

$$ \frac{250}{25}=10 $$

Hence, the number of rungs is

$$ 10+1=11 $$

The lengths of the rungs form an Arithmetic Progression (AP):

$$ 45,\;43,\;41,\;\ldots,\;25 $$

Here,

$$ a=45,\qquad l=25,\qquad n=11 $$

The common difference is

$$ d=\frac{25-45}{11-1} =\frac{-20}{10} =-2 $$

The total length of wood required is the sum of the lengths of all the rungs.

Using the formula,

$$ S_n=\frac{n}{2}(a+l) $$

Substituting the values,

$$ S_{11}=\frac{11}{2}(45+25) $$

$$ =\frac{11}{2}\times70 $$

$$ =11\times35 $$

$$ =385\text{ cm} $$

Converting into metres,

$$ 385\text{ cm}=3.85\text{ m} $$

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Answer

The total length of wood required for the rungs is

$$ \boxed{385\text{ cm}} $$

or

$$ \boxed{3.85\text{ m}} $$

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Question 4

Question:

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 $x$.

Solution:

The sum of the numbers preceding house $x$ is

$$ S_{x-1}=\frac{(x-1)x}{2} $$

The sum of the numbers from 1 to 49 is

$$ S_{49}=\frac{49\times50}{2}=1225 $$

The sum of the numbers following house $x$ is

$$ S_{49}-S_x $$

According to the question,

$$ S_{x-1}=S_{49}-S_x $$

Substituting the formulas,

$$ \frac{x(x-1)}{2}=1225-\frac{x(x+1)}{2} $$

Multiplying both sides by 2,

$$ x(x-1)=2450-x(x+1) $$

$$ x^2-x=2450-x^2-x $$

$$ 2x^2=2450 $$

$$ x^2=1225 $$

$$ x=35 $$

Verification:

Sum of the house numbers before 35:

$$ \frac{34\times35}{2}=595 $$

Sum of the house numbers after 35:

$$ 1225-\frac{35\times36}{2} $$

$$ =1225-630 $$

$$ =595 $$

Both sums are equal.

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Answer

The required house number is

$$ \boxed{x=35} $$

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Question 5

Question:

A small terrace at a football ground comprises 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of $$ \frac14\text{ m} $$ and a tread of $$ \frac12\text{ m}. $$ Calculate the total volume of concrete required to build the terrace.

Solution:

Length of each step

$$ =50\text{ m} $$

Rise of each step

$$ =\frac14\text{ m} $$

Tread of each step

$$ =\frac12\text{ m} $$

The volume of the first step is

$$ \frac14\times\frac12\times50 =\frac{25}{4}\text{ m}^3 $$

The second step has twice the height, so its volume is

$$ 2\times\frac{25}{4} $$

Similarly, the volumes of the steps form an Arithmetic Progression:

$$ \frac{25}{4},\; 2\left(\frac{25}{4}\right),\; 3\left(\frac{25}{4}\right),\; \ldots,\; 15\left(\frac{25}{4}\right) $$

Here,

$$ a=\frac{25}{4},\qquad d=\frac{25}{4},\qquad n=15 $$

The total volume is

$$ S_{15} =\frac{15}{2}\left[2\left(\frac{25}{4}\right)+14\left(\frac{25}{4}\right)\right] $$

$$ =\frac{15}{2}\times16\times\frac{25}{4} $$

$$ =15\times2\times25 $$

$$ =750\text{ m}^3 $$

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Answer

The total volume of concrete required is

$$ \boxed{750\text{ m}^3} $$