Arithmetic Progressions | Exercise 5.4

Question 3

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 212 m2\dfrac{1}{2}\text{ m} apart, what is the length of the wood required for the rungs?

[Hint : Number of rungs = 25025+1\dfrac{250}{25} + 1]

Question diagram 1
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Solution
Understand the Question
  • The total distance between the top and bottom rungs is 212 m=250 cm2\dfrac{1}{2}\text{ m} = 250\text{ cm}, and the rungs are spaced 25 cm25\text{ cm} apart.
  • The number of rungs is given by Total distanceDistance between rungs+1\dfrac{\text{Total distance}}{\text{Distance between rungs}} + 1.
  • Since the lengths of the rungs decrease uniformly, they form an Arithmetic Progression (AP) where:
    • First term (aa) = length of top rung = 25 cm25\text{ cm}
    • Last term (ll or ana_n) = length of bottom rung = 45 cm45\text{ cm}
  • The total length of wood required is the sum of the lengths of all rungs (SnS_n).

Step 1 · Calculate the Number of Rungs

Total distance between the top and bottom rungs: Total distance=212 m=2.5×100=250 cm\text{Total distance} = 2\dfrac{1}{2}\text{ m} = 2.5 \times 100 = 250\text{ cm}

Distance between successive rungs =25 cm= 25\text{ cm}.Diagram 1

Number of gaps=25025=10\text{Number of gaps} = \dfrac{250}{25} = 10

Number of rungs (n)=Number of gaps+1=10+1=11\begin{aligned} \text{Number of rungs } (n) &= \text{Number of gaps} + 1 \\[0.6em] &= 10 + 1 \\[0.6em] &= 11 \end{aligned}

Step 2 · Define the Arithmetic Progression

Let the lengths of the rungs form an AP from top to bottom:

  • First term (aa) =25 cm= 25\text{ cm}
  • 11th11^{\text{th}} term (a11a_{11}) =45 cm= 45\text{ cm}
  • Number of terms (nn) =11= 11

Using an=a+(n1)da_n = a + (n - 1)d

a11=a+(111)d45=25+10d10d=20d=2 cm\begin{aligned} a_{11} &= a + (11 - 1)d \\[0.6em] 45 &= 25 + 10d \\[0.6em] 10d &= 20 \\[0.6em] d &= 2\text{ cm} \end{aligned}

Step 3 · Calculate the Total Length of Wood

Using the sum formula Sn=n2(2a+(n1)d)S_n = \dfrac{n}{2}(2a + (n - 1)d) with n=11n = 11, a=25a = 25, and d=2d = 2

S11=112(2(25)+(111)2)=112(50+20)=112(70)=11×35=385 cm\begin{aligned} S_{11} &= \dfrac{11}{2}(2(25) + (11 - 1)2) \\[0.6em] &= \dfrac{11}{2}(50 + 20) \\[0.6em] &= \dfrac{11}{2}(70) \\[0.6em] &= 11 \times 35 \\[0.6em] &= 385\text{ cm} \end{aligned}
Answer

385 cm385\text{ cm}

Common Mistakes
  • Fencepost Error: Forgetting to add 11 when finding the number of rungs (dividing 25025=10\dfrac{250}{25} = 10 gives the number of spaces between rungs, so the number of rungs is 10+1=1110 + 1 = 11).
  • Unit Mismatch: Not converting 212 m2\dfrac{1}{2}\text{ m} to cm\text{cm} before dividing by 25 cm25\text{ cm}.

More questions in Exercise 5.4

Q1

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

[Hint : Find nn for an<0a_n < 0]

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.

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 212 m2\dfrac{1}{2}\text{ m} apart, what is the length of the wood required for the rungs?

[Hint : Number of rungs = 25025+1\dfrac{250}{25} + 1]

Q4

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

[Hint : Sx1=S49SxS_{x-1} = S_{49} - S_x]

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 14 m\dfrac{1}{4} \text{ m} and a tread of 12 m\dfrac{1}{2} \text{ 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 = 14×12×50 m3\dfrac{1}{4} \times \dfrac{1}{2} \times 50 \text{ m}^3]

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