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Arithmetic Progressions
RD Sharma
CBSE
Grade 10

Question:

A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii $0.5\text{ cm}, 1.0\text{ cm}, 1.5\text{ cm}, 2.0\text{ cm}, \dots$. What is the total length of such a spiral made up of $13$ consecutive semicircles? (Take $\pi = 22/7$)

Step-by-Step Solution

Key Concept: Length of $i^{\text{th}}$ semicircle $= l_i = \pi r_i = \pi(0.5 i)$. Total length $= \pi(0.5)[1 + 2 + 3 + \dots + 13] = \pi(0.5) \dfrac{13 \times 14}{2} = \pi(0.5)(91) = \dfrac{22}{7} \times 0.5 \times 91 = 22 \times 0.5 \times 13 = 11 \times 13 = 143\text{ cm}$.
Lengths $l_1 = 0.5\pi, l_2 = 1.0\pi, l_3 = 1.5\pi, \dots, l_{13} = 6.5\pi$. Form an A.P. with $a = 0.5\pi, d = 0.5\pi, n = 13$. [1.5 Marks]
Total length $= \dfrac{13}{2}[l_1 + l_{13}] = \dfrac{13}{2}[0.5\pi + 6.5\pi] = \dfrac{13}{2}[7\pi] = \dfrac{91}{2}\pi$. [1.5 Marks]
Total length $= \dfrac{91}{2} \times \dfrac{22}{7} = 13 \times 11 = 143\text{ cm}$. [2.0 Marks]

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🎯 Official CBSE Marking Scheme:
Forming A.P. of semicircle lengths: 1.5 Marks
Evaluating total sum in terms of $\pi$: 1.5 Marks
Substituting $\pi = 22/7$ to get $143\text{ cm}$: 2.0 Marks

Correct Answer:
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