Makayla Reilly
2022-09-17
Answered

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

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soyafh

Answered 2022-09-18
Author has **17** answers

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

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

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

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

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

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