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 , …
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 Tn = a + (n - 1)d
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.
The difference between two consecutive term in arithmetic progression is the same.
So, T2 - T1 = 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 9th term of the progression. Find the value of each of the three terms.
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