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Chapter 22: Problem 662

If the first term of an arithmetic progression is 7, and the common difference is \(-2\), find the fifteenth term and the sum of the first fifteen terms.

Short Answer

Expert verified
The 15th term of the arithmetic progression is \(T_{15} = -21\) and the sum of the first 15 terms is \(S_{15} = -105\).

Step by step solution

01

Find the 15th term

Using the formula for the nth term of an arithmetic progression, we can find the 15th term: \(T_{15} = a + (15-1)d\) Given the first term, a = 7, and the common difference, d = -2, we plug them into the formula: \(T_{15} = 7 + (14)(-2)\)
02

Calculate the 15th term

Now, perform the calculations: \(T_{15} = 7 - 28\) \(T_{15} = -21\) Now we know that the 15th term of the arithmetic progression is -21.
03

Calculate the sum of the first 15 terms

We will use the formula for the sum of the first n terms of an arithmetic progression to find the sum of the first 15 terms: \(S_{15} = \frac{15}{2}(2a + (15-1)d)\) Given the first term, a = 7, and the common difference, d = -2, we plug them into the formula: \(S_{15} = \frac{15}{2}(2(7) + (14)(-2))\)
04

Calculate the sum

Now, perform the calculations: \(S_{15} = \frac{15}{2}(14 - 28)\) \(S_{15} = \frac{15}{2}(-14)\) \(S_{15} = -15 \cdot 7\) \(S_{15} = -105\) Now we know that the sum of the first 15 terms of the arithmetic progression is -105. So, the 15th term of the arithmetic progression is -21, and the sum of the first 15 terms is -105.

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Most popular questions from this chapter

Find the sum of the geometric series \(30+10+3(1 / 3)+\ldots\) \(+30(1 / 3)^{\mathrm{n}-1}+\ldots\)

The sum of three numbers in arithmetic progression is 27 , and the sum of their squares is 293 ; find them.

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The sum of an infinite number of terms in geometric progression is 15 , and the sum of their squares is 45 ; find the sequence. Assume that the common ratio of the G.P. is less than 1 .

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