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Chapter 22: Problem 676

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

Short Answer

Expert verified
The three numbers in the arithmetic progression are 4, 9, and 14.

Step by step solution

01

Assign variables to the given information

Let the three numbers be \(a - d\), \(a\), and \(a + d\), where \(a\) is the middle number and \(d\) is the common difference between the consecutive terms of the arithmetic progression.
02

Write equations based on the given information

We are given that the sum of these three numbers is 27, so we can write the equation: \((a - d) + a + (a + d) = 27\) We are also given that the sum of their squares is 293, so we can write the equation: \((a - d)^2 + a^2 + (a + d)^2 = 293\)
03

Simplify the first equation

Simplify the equation \((a - d) + a + (a + d) = 27\) \(3a = 27\) Now, we can solve for \(a\): \(a = 9\)
04

Substitute \(a\) in the second equation

Substitute \(a = 9\) in the equation \((a - d)^2 + a^2 + (a + d)^2 = 293\): \((9 - d)^2 + 9^2 + (9 + d)^2 = 293\)
05

Simplify and solve for \(d\)

Simplify the equation \((9 - d)^2 + 9^2 + (9 + d)^2 = 293\): \((81 - 18d + d^2) + 81 + (81 + 18d + d^2) = 293\) Combine like terms: \(2d^2 + 243 = 293\) Subtract 243 from both sides: \(2d^2 = 50\) Divide both sides by 2: \(d^2 = 25\) Now, we can solve for \(d\): \(d = 5\) or \(d = -5\)
06

Calculate the three numbers

We found \(a = 9\) and two possible values of \(d\): \(5\) and \(-5\). If \(d = 5\), the three numbers are: \(a - d = 9 - 5 = 4\) \(a = 9\) \(a + d = 9 + 5 = 14\) If \(d = -5\), the three numbers are: \(a - d = 9 - (-5) = 14\) \(a = 9\) \(a + d = 9 + (-5) = 4\) Both cases give us the same set of numbers but in different orders: 4, 9, and 14. These are the three numbers in the arithmetic progression.

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Most popular questions from this chapter

Find the 9 th term of the harmonic progression \(3,2,3 / 2, \ldots .\)

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 .

The first term of a geometric progression is 27, the nth term is \(32 / 9\), and the sum of n terms is \(665 / 9\). Find \(\mathrm{n}\) and \(\mathrm{r}\).

Find the next three terms of the geometric progression \(27,-9,3,-1 \ldots\)

Find the first six terms of the sequence determined by the function \(g(x)\), where \(x=1,2,3,4,5,6\) \(g(x)=x^{2} / x !, x\) a positive integer
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