Understanding the Sum of an Arithmetic Progression

An arithmetic progression is a sequence of numbers in which each term after the first is obtained by adding a constant, called the common difference, to the previous term. This article will guide you through a detailed example to find the sum of an arithmetic progression with 30 terms. We will explore the given conditions and apply the relevant formulas to solve the problem.

Given Conditions and Solution

The problem statement provides the following details:

The sum of the first 10 terms of an arithmetic progression is 185. The 21st term is 15 more than the 16th term. Utilize the arithmetic progression formulas to find the sum of the 30 terms.

Step-by-Step Solution

Let's denote the first term as a and the common difference as d. We will use the arithmetic progression formulas to solve the problem.

Given that the sum of the first 10 terms is 185:

S10 185

The formula for the sum of the first n terms of an arithmetic progression is:

Sn n/2(2a (n-1)d)

Substituting n 10:

185 10 / 2 (2a 9d)

37 2a 9d

Given that the 21st term is 15 more than the 16th term:

T21 T16 15

The formula for the nth term of an arithmetic progression is:

Tn a (n-1)d

T21 a 20d

T16 a 15d

Substituting the given condition:

a 20d a 15d 15

20d - 15d 15

5d 15

d 15/5 3

Substitute d 3 into the equation 37 2a 9d:

37 2a 27

10 2a

a 10/2 5

Now that we have the first term a 5 and the common difference d 3, we can find the sum of the 30 terms:

S30 30/2(2a 29d)

S30 15 (10 87)

S30 15 × 97

S30 1455

Conclusion

The sum of the 30 terms of the arithmetic progression is 1455. By applying the formulas for the sum of an arithmetic progression and the nth term, we were able to solve the given problem step by step.

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