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 cm, 1.5 cm, 2.0 cm, ……… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take \(\pi = \frac{22}{7} \) )

We know,
Perimeter of a semi-circle = \(\pi\)r

Therefore,
\(P_1 = \pi(0.5) = {{\pi} \over {2}}\) cm
\(P_2 = \pi(1) = \pi\) cm
\(P_3 = \pi(1.5) = {{3pi} \over { 2}}\) cm

Where, \(P_1, P_2, P_3\) are the lengths of the semi-circles.

Hence we got a series here, as,

\({{\pi} \over {2}} , \pi, {{3\pi} \over { 2}}, 2\pi,\) ….

\(P_1 = {{\pi} \over {2}}\) cm
\(P_2 = \pi\) cm
Common difference, \(d = P_2- P_1 = \pi – {{\pi} \over {2}} = {{\pi} \over {2}}\)
First term = \(P_1= a = {{\pi} \over {2}}\) cm

By the sum of n term formula, we know,
\(S_n = {{n} \over {2}} [2a + (n – 1)d]\)

Therefor, Sum of the length of 13 consecutive circles is;

\(S_{13} = {{13} \over {2}} [2({{\pi} \over {2}}) + (13 – 1){{\pi} \over {2}}]\)
= \( \frac{13}{2} [\pi + 6\pi]\)
=\(\frac{13}{2} (7\pi)\)
= \(\frac{13}{2} × 7 × \frac{22}{7}\)
= \(143\) cm