(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 = \(\frac{-9}{4}, \frac{3}{4}, \frac{15}{4}, \frac{27}{4}, \frac{39}{4}\)