Elaborate the practical examples of ARITHMETIC PROGRESSION, GEOMETRIC PROGRESSION?i want only few examples.

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You need to remember that between any two consecutive terms of an arithmetic progression exists the following relation such that:

`a_n = a_(n-1) + d`

`a_(n-1) ` and `a_n`  represent two consecutive terms of an arithmetic progression

d represents a constant called common difference

You may also use the following formula that allows you to find any term of arithmetic progression, if you know the first term and the common difference, such that:

`a_n = a_1 + (n-1)*d`

Considering as a worked example the following problem that provides the first terms of an arithmetic progression, `a_1 = 2`  and the third term `a_3 = 8`  and it requests you to find the 6th term `a_6` , you need to perform the following steps:

`a_3 = a_1 + 2d`  (finding the common difference)

`8 = 2 + 2d => 2d = 6 => d = 3`

`a_6 = a_1 + 5d`  (finding the term `a_6` )

`a_6 = 2 + 5*3 => a_6 = 17`

Hence, evaluating the common difference `d`  and the term `a_6` , for the considered arithmetic progression, using the formula `a_n = a_1 + (n-1)*d` , yields `a_6 = 17.`

You should also remember that between any two consecutive terms of a geometric progression exists the following relation such that:

`a_n = a_(n-1)*q`

`a_(n-1) `  and `a_n`  represent two consecutive terms of an arithmetic progression

q represents a constant called common ratio

`a_n = a_1*q^(n-1)`

Considering as a worked example the following problem that provides the first terms of a geometric progression, `a_1 =3 ` and the second term `a_2 =9`  and it requests you to find the term `a_10` , you need to perform the following steps:

`a_2 = a_1*q`  (finding the common ratio)

`9 = 3*q => q = 3`

`a_10 = a_1*q^(n-1) ` (finding the term `a_10` )

`a_10 = 3*3^(10-1) => a_10 = 3^(1+10-1) = 3^10`

Hence, evaluating the common ratio q and the term `a_10` , for the considered geometric progression, using the formula `a_n = a_1*q^(n-1), ` yields `a_10 = 3^10.`

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