NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 18
NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 18
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 cm, 1.0 cm, 1.5 cm, 2.0 cm,... What is the total length of such a spiral made up of thirteen consecutive 22 semicircles? (Take pi = 22/7)
Given:
Spiral is made of semicircles.
Total number of semicircles = 13.
Radii of semicircles are 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ...
π = 22/7
To Find:
Total length of the spiral.
Formula:
Length of a semicircle = πr
The nth term of an arithmetic progression is given by an = a + (n-1)d.
The sum of an arithmetic progression with n terms is given by Sn = n2(2a + (n-1)d).
Solution:
The radii of the semicircles are in an arithmetic progression with
first term a = 0.5 and
common difference d = 0.5.
The nth term of an arithmetic progression is given by an = a + (n-1)d.
Therefore, the radius of the 13th semicircle is
a13 = 0.5 + (13-1)0.5
⇒ a13 = 6.5 cm.
The sum of an arithmetic series is given by Sn = n2(2a + (n-1)d).
Therefore, the sum of the radii is
⇒ S13 = 132 (2(0.5) + (13-1)0.5)
⇒ S13 = 45.5 cm.
Total length of the spiral = π(sum of radii)
⇒ Total length of the spiral = 227 × 45.5
⇒ Total length of the spiral = 143 cm.
Result:
The total length of the spiral is 143 cm.
Next question solution:
NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 19
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