Design Pattern Question (Math Help)

Introduction to Design pattern question:

A Pattern defines a group of numbers in which all the numbers are related with each by a specific rule. A pattern is the process of multiplying the previous time by a stable factor. Such a progression is called design patterns. Patterns give us enormous joy to find the connection between the numbers which different number forms of design patterns. The stable factor is called common ratio (C.R) in patterns. And now we can se about to design some pattern question.

Design pattern question in arithmetic progression:

Question 1:

Is the sequence 10, 4, -2, -8, ... an A.P.

Solution:

In the given sequence we find 4 - 10 = -2 - 4 = -8 - (-2) = - 6

The common difference is -6. Hence the given sequence is an A.P.

Question 2:

Is the sequence described by an = 2n^2 + 1 an A.P.?

Solution:

an = 2n^2 + 1

a1= 2(1)^2 + 1 = 3, a^2 = 2(2)^2 + 1 = 9

a3= 2(3)^2 + 1 = 19, a4 = 2(4)^2 + 1 = 33

The sequence is 3, 9, 19, 33, ...

Here, 9 - 3 = 6

19 - 9 = 10

33 - 19 = 14

The difference is not the same.

The given sequence is not an A.P.

Question 3:

Write down the A.P. and its general term if a = 3, d = 7.

Solution:

Consider the A.P in the form a, a + d, a + 2d.

? The A.P is 3, 3 + 7, 3 + 14, ... or 3, 10, 17...

General term tn = a (n - 1)d = 3 + (n - 1) 7 = 7n - 4

Question 4:

Find 4 numbers between 3 and 38 which are in an A.P.

Solution:

Consider the A.P in the form a, a + d, a + 2d, ...

Here a = 3, and a + 5d = 38

'=>' 5d = 35, '=>' d = 7

? The A.P. is 3, 10, 17, 24, 31, 38...

? The 4 numbers between 3 and 38 are 10, 17, 24, 31.

Design pattern question in Geometric progression:

Question 1:

Find three numbers in G.P such that their sum is 7 and the sum of the

reciprocals is 7/4.

Solution:

Let the three numbers in G.P be a, ar, ar^2

Sum of the numbers = a + ar + ar^2 = 7, a (1 + r + r^2) = 7 (1)

Sum of the reciprocals = 1/a + 1/ar + 1/ar^2

1 + r + r^2
= _________ (2)
ar^2

Dividing (1) and (2) we get (ar)^2 = 4, ar = + 2 '=>' a = + 2/r

Substituting a = 2/r in (1) we get

2/r (1 + r + r^2) = 7 '=>' 2 (1 + r + r^2) = 7r

'=>' 2r2 - 5r + 2 = 0 '=>'r = 1/2 or 2

If r = 1/2 then a = 4. ? The numbers are 4, 2, 1...

If r = 2 then a = 1 ? The numbers are 1, 2, 4

Question 2:

If a, b, c, d are in G.P show that (a - b + c) (b + c + d) = ab + bc + cd

Solution:

a, b, c, d are in G.P. '=>' b = ar, c = ar^2, d = ar3

LHS = (a - b + c) (b + c + d) = (a - ar + ar^2) (ar + ar2 + ar3)

= a2 (1 - r + r^2) (r + r^2 + r3)

= a2r (1 - r + r^2) (1 + r + r^2)

= a2r (1 + r^2 + r4)

= a^2r + a^2r3 + a^2r5

= a(ar) + (ar) (ar^2) + (ar^2) (ar^3)

= ab + bc + cd = RHS

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