# Ex.5.3 Q18 Arithmetic Progressions Solution - NCERT Maths Class 10

## Question

A spiral is made up of successive semicircles, with centres alternately at \(A\) and \(B,\) starting with centre at \(A\) of radii \(0.5,\) \(1.0 \,\rm{cm}, 1.5 \,\rm{cm}, 2.0 \,\rm{cm}, ......... \)as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles?

\(\begin{align}\left[ {{\rm{Take}}\,\,\pi = \frac{{22}}{7}} \right] \end{align}\)

## Text Solution

**What is Known?**

Radii of the \(13\) semicircles.

**What is Unknown?**

Length of the spiral made.

**Reasoning:**

Sum of the first \(n\)terms of an AP is given by \({S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\)

Where \(a\) is the first term, \(d\)is the common difference and \(n\) is the number of terms.

**Steps:**

Semi-perimeter of circle, \(l = \pi r\)

\[\begin{align}{{l}_{1}}& =\pi \times \left( 0.5\,\,\text{cm} \right)=0.5\pi \,\,\text{cm } \\ {{l}_{2}}&=\pi \times \left( 1\,\,\text{cm} \right)=\pi \,\,\text{cm} \\ {{l}_{3}} &=\pi \times \left( 1.5\,\,\text{cm} \right)=1.5\pi \,\text{cm} \\\end{align}\]

Therefore, \({l_1},{\rm{ }}{l_2},{\rm{ }}{l_3},\) i.e. the lengths of the semi-circles are in an A.P.,

\[\begin{align}&0.5\pi ,\pi ,1.5\pi ,2\pi\dots\dots\dots\\&a = 0.5\pi \\&d = \pi - 0.5\pi = 0.5\pi\end{align}\]

We know that the sum of \(n\) terms of an A.P. is given by

\[\begin{align}{S_n} &= \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\\{S_{13}} &\!=\!\frac{{13}}{2}\left[ {2 \!\times\! \left( {0.5\pi } \right)\!+\!\left( {13\!-\!1} \right)\left( {0.5\pi } \right)}\!\right]\\ &= \frac{{13}}{2}\left[ {\pi + 6\pi } \right]\\ &= \frac{{13}}{2} \times 7\pi \\ &= \frac{{13}}{2} \times 7 \times \frac{{22}}{7}\\ &= 143\end{align}\]

Therefore, the length of such spiral of thirteen consecutive semi-circles will be \(143\, \rm{cm.}\)