Exercise 5.4 — Solutions

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Q1. Which term of the AP $121,117,113,\dots$

 is its first negative term?

Given

$a=121$

, common difference

$d=117-121=-4$

General term:

$$a_n=a+(n-1)d=121-4(n-1)\mathrm{.}$$

Find smallest integer

$n$

with

$a_n<0$

:

$$121-4(n-1)<0\Rightarrow 121-4n+4<0\Rightarrow 125<4n\Rightarrow n>\mathrm{31.25.}$$

Hence the first integer

$n$

is

$32$

.
Check:

$$a_{31}=121-4(30)=121-120=1(>0),a_{32}=121-4(31)=121-124=-3(<0)\mathrm{.}$$

Answer: the 32nd term is the first negative term (value

$-3$

).

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Q2. The sum of the 3rd and 7th terms of an AP is $6$

and their product is $8$

. Find the sum of the first 16 terms.

Let

$a$

a be the first term and

$d$

d the common difference. Then

$$a_3=a+2d,a_7=a+6d\mathrm{.}$$

Given:

$$\begin{array}{cccc}& (a+2d)+(a+6d)=6\Rightarrow 2a+8d=6\Rightarrow a+4d=3.& & \text{(1)}\end{array}$$

And

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

Set

$x=a+4d$

. Then

$a+2d=x-2d,a+6d=x+2d$

. Their product:

$$(x-2d)(x+2d)=x^2-4d^2=8.$$

From (1)

$x=3$

. So

$$3^2-4d^2=8\Rightarrow 9-4d^2=8\Rightarrow 4d^2=1\Rightarrow d=\pm \frac{1}{2}\mathrm{.}$$

Find

$a$

from

$a+4d=3$

• If

  $d=\frac{1}{2}$​:

  $a=3-4\cdot \frac{1}{2}=3-2=1.$

• If

  $d=-\frac{1}{2}$

  ​:

  $a=3-4(-\frac{1}{2})=3+2=5.$

Now

$S_{16}={\displaystyle \frac{16}{2}}(2a+15d)=8(2a+15d)$

• For

  $a=1,d=\frac{1}{2}$

  $$2a+15d=2+7.5=9.5,S_{16}=8\times 9.5=76.$$

• For

  $a=5,d=-\frac{1}{2}$

  $$2a+15d=10-7.5=2.5,S_{16}=8\times 2.5=20.$$

Both

$(a,d)=(1,\frac{1}{2})$

and

$(5,-\frac{1}{2})$

satisfy the given conditions (they just reverse the order of the two specified terms).

Answer: there are two possible sums depending on the AP:

$$\boxed{{\displaystyle S_{16}=76\text{ or }{S}_{16}=20.}}$$

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Q3. Ladder rungs 25 cm apart. Rungs decrease uniformly in length from $45$

 cm (bottom) to $25$

 cm (top). If the top and bottom rungs are $2\frac{1}{2}$

​ m apart, what is the total length of wood required for the rungs?

(Hint in book:

$\text{number of rungs}={\displaystyle \frac{250}{25}}+\mathrm{1 }$

number of rungs=25250​+1 — i.e. distance 250 cm.)

Interpretation / data

• Distance between top and bottom rungs

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

• Spacing between successive rungs

  $=25\text{ cm}$.

• Number of rungs:

  $$n=\frac{250}{25}+1=10+1=11.$$

• Lengths form an AP with first term

  $a=\mathrm{45 }$cm, last term

  $l=25$ cm,

  $n=11$.

Sum of lengths:

$$\text{Total length}=S_{11}=\frac{11}{2}(a+l)=\frac{11}{2}(45+25)=\frac{11}{2}\cdot 70=11\cdot 35=385\text{ cm}\mathrm{.}$$

Convert to metres:

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

Answer: total wood required

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

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Q4. Houses numbered $1$

to $49$

. Show there is a value $x$

such that the sum of the numbers of houses preceding house $x$

equals the sum of the numbers following it. Find $x$

.

Let

$S_k$

Sk​ denote the sum

$1+2+\cdots +k={\displaystyle \frac{k(k+1)}{2}}$

We want:

$$S_{x-1}=S_{49}-S_x\mathrm{.}$$

Compute:

$$\frac{(x-1)x}{2}=\frac{49\cdot 50}{2}-\frac{x(x+1)}{2}\mathrm{.}$$

Multiply by

$2$

2 and simplify:

$$x(x-1)=2450-x(x+1)\Rightarrow 2x^2-2450=0.$$

So

$x^2=1225$

and

$x=35$

(positive root).

Check:

$$S_{34}=\frac{34\cdot 35}{2}=595,S_{49}-S_{35}=1225-630=595.$$

Answer:

$x=35$

. The sums on both sides are

$595$

.

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Q5. A terrace has 15 steps, each $50$

 m long (along the step). Each step has rise $=\frac{1}{4}$

 m and tread $=\frac{1}{2}$

​ m. Calculate total volume of concrete required.

Volume of one step (rectangular block)

$=\text{length}\times \text{tread}\times \text{rise}$

$$V_1=50\times \frac{1}{2}\times \frac{1}{4}=50\times \frac{1}{8}=6.25{\text{m}}^3\mathrm{.}$$

Assuming all 15 steps are solid (each identical in cross-section), total volume:

$$V_{\text{total}}=15\times 6.25=93.75{\text{m}}^3\mathrm{.}$$

Answer: total concrete required

$=93.75{\text{m}}^3\mathrm{}$