**.**

*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, T_{2 }- T_{1 = }T_{3} - T_{2}

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} = *x*^{2} + 16*x**x*^{2} + 4*x* + 4 = *x*^{2} + 16*x*

4 = 12*x**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 *+* *2*d*

Since the last of these terms, *x + 2d*, is 6 less than T_{9}, so

T_{9} – (*x *+* *2*d*) = 6

*a* + 8*d* – *x* – 2*d* = 6

2 + 8*d* – *x* – 2*d* = 6

6*d* – *x* = 4 ………………… ( i )

*x* + *x* + *d* + *x* + 2*d* = 51

*x* + 3*d* = 51

*x* + *d* = 17 ………………… ( ii )

*d* = 21

*d* = 3, *x* = 14

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