[Add Maths Form 5] d For Difference

An arithmetic progression/sequence is a sequence of numbers in which the difference between two successive number is a constant, d.

Thus, the form of an arithmetic sequence is a , a + d , a + 2d , a + 3d , …

Eg:
2, 5, 8, 11, 14, … is an arithmetic progression/sequence with common difference 3.

To find the n-th term in an arithmetic sequence, use T_n = a + (n - 1)d

Example:

1) Find the value of x, in which lg x, lg(x + 2) and lg(x + 16) are three consecutive terms of an arithmetic progression.

Solution:
The difference between two consecutive term in arithmetic progression is the same.

So, T₂ - T_{1 =} T₃ - T₂
lg(x + 2) - lg x = lg(x + 16) - lg(x + 2)
lg( (x + 2) / x ) = lg( (x + 16) / (x + 2) )
(x + 2) / x = (x + 16) / (x + 2)
(x + 2) ² = x² + 16x
x² + 4x + 4 = x² + 16x
4 = 12x
x = 1/3

2) An arithmetic progression whose first term is 2 includes three consecutive terms that have a sum of 51. The last of these terms is 6 less than the 9^th term of the progression. Find the value of each of the three terms.

Solution:

Let a = 2 be the first term

Then, write the arithmetic progression as: 2, ….x, x + d, x + 2d

Since the last of these terms, x + 2d, is 6 less than T₉, so

T₉ – (x + 2d) = 6

a + 8d – x – 2d = 6

2 + 8d – x – 2d = 6

6d – x = 4 ………………… ( i )

Also, since x + x + d + x + 2d = 51

3x + 3d = 51

x + d = 17 ………………… ( ii )

( i ) + ( ii )

7d = 21

d = 3, x = 14

Hence, the three terms are 14, 17 and 20 respectively