The first three terms of an arithmetic sequence are 2k+3, 5k-2 and 10k-15, how do you show that k =4?

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The first three terms of an arithmetic sequence are 2k+3, 5k-2 and 10k-15, how do you show that k =4?

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Answer 1 · 2016-05-17T02:54:41

Use the definition of an arithmetic sequence to set up a system of two equations and two unknowns and solve.

Explanation:

Since the sequence is arithmetic, there is a number $d$ $d$ (the "common difference") with the property that $5 k - 2 = (2 k + 3) + d$ $5k-2=(2k+3)+d$ and $10 k - 15 = (5 k - 2) + d$ $10k-15=(5k-2)+d$ . The first equation can be simplified to $3 k - d = 5$ $3k-d=5$ and the second to $5 k - d = 13$ $5k-d=13$ . You can now subtract the first of these last two equations from the second to get $2 k = 8$ $2k=8$ , implying that $k = 4$ $k=4$ .

Alternatively, you could have set $d = 3 k - 5 = 5 k - 13$ $d=3k-5=5k-13$ and solved for $k = 4$ $k=4$ that way instead of subtracting one equation from the other.

It's not necessary to find, but the common difference $d = 3 k - 5 = 3 \cdot 4 - 5 = 7$ $d=3k-5=3*4-5=7$ . The three terms in the sequence are 11, 18, 25.

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The first three terms of an arithmetic sequence are 2k+3, 5k-2 and 10k-15, how do you show that k =4?

1 Answer

Explanation:

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