<PRE><B>Question 1203960</B>


In a certain spider web, the lengths of the shortest and longest threads are 3.4mm and 50.2mm respectively. The successive equally spaced threads are 1.2mm apart. If the length of the cross threads form an Arithmetic Progression (A.P), find the number of cross threads.

 Length of the shortest  thread are 3.4mm

Length of the longest  thread are 50.2mm

The successive equally spaced threads are 1.2mm apart

 length of the cross threads form an Arithmetic Progression

Here a 3.4 , tn = 50.2  d= 1.2

tn = a+(n-1)*d

50.2= 3.4+ (n-1)* 1.2

50.2 - 3.4 = (n-1)*1.2

46.8= (n-1) *1.2

46.8/1.2= n-1

39 = n-1

n=40

 The number of cross threads =40


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