(a) How many numbers between 75 and 500 are divisible by 7?
(b) The 8th term of an Arithmetic Progression (A.P) is 5 times the 3rd term while the 7th term is 9 greater than the 4th term. Write the first 5 terms of the A.P.
Explanation
(a) In the set of numbers from 75, 76, 77, … , 497, 498, 499, 500, you will find that the first term which is the first number divisible by 7 = 77 and the last term = 497.
Given the first and last terms of an A.P, you can use the formula
(n = frac{l – a}{d} + 1) to find the number of terms in the sequence.
First term = a = 77
Last term = l = 497
Common difference = d = 7
(n = frac{497 – 77}{7} + 1)
(n = frac{420}{7} + 1)
= (60 + 1)
= 61.
There are 61 numbers between 75 and 500 which are divisible by 7.
(b) (T_{n} = a + (n – 1)d) (for an A.P)
(T_{8} = a + 7d)
(T_{3} = a + 2d)
(implies a + 7d = 5(a + 2d))
(T_{7} = a + 6d)
(T_{4} = a + 3d)
(implies a + 6d = a + 3d + 9)
(a + 7d = 5a + 10d implies 5a – a + 10d – 7d = 0)
(4a + 3d = 0 … (1))
(a + 6d = a + 3d + 9 implies a – a + 6d – 3d = 9)
(3d = 9 implies d = 3)
Putting d = 3 in (1),
(4a + 3(3) = 0 implies 4a + 9 = 0)
(4a = -9 implies a = frac{-9}{4})
(therefore) The second term of the sequence = (frac{-9}{4} + 3 = frac{3}{4})
Third term = (frac{3}{4} + 3 = frac{15}{4})
Fourth term = (frac{15}{4} + 3 = frac{27}{4})
Fifth term = (frac{27}{4} + 3 = frac{39}{4})
The first 5 terms of the sequence = (frac{-9}{4}, frac{3}{4}, frac{15}{4}, frac{27}{4}, frac{39}{4})