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For what value of k will the consecutive terms 2k 1 3k 3 and 5k 1 form an arithmetic progression?

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For what value of k will the consecutive terms 2k 1 3k 3 and 5k 1 form an arithmetic progression?

The given terms are 2k + 1, 3k + 3 and 5k − 1. Hence, the value of k for which the given terms are in an AP is 6.

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What is the value of K if 2k 1 2k 1 3k 4 are in an arithmetic progression?

Answer: The value of k is 3.

For what value of K Will K 9 2k 1 and 2k 7 are the consecutive terms of an AP?

=18
If k+9,2k-1 and 2k+7 are the consecutive terms of an A.P, then the common difference ‘d’ will be same. Hence, if k=18, the consecutive terms k+9,2k-1 and 2k+7 are in AP.

For what value of k will the terms 2k 1/8 3k form an AP?

we know that to form an Ap the common differences will be same. =>k=3. so the value of k will be 3.

For what value of k will consecutive terms 2k 1?

6 is the value of k for the consecutive terms 2k+1, 3k+3, 5k-1 which forms an AP. k = 6. Therefore, 6 is the value of k for the given consecutive terms.

For what value of k the numbers K 2 4k 1 and 5k +2 will be the consecutive terms of an AP?

Solution:- We know, The common difference between all the terms in an A.P. is always equal. Therefore the value of k is 1.

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For what value of K Will K 2k 1 and 2k 1 are the consecutive terms of an AP?

It is given that k,(2k -1)and (2k +1) are the three successive terms of an AP. Hence, the value of k is 3.

For what value of K K 2 4k 6 3k 2 are three consecutive terms of AP?

It is given that (3k -2) ,(4k -6) and (k +2) are three consecutive terms of an AP. Hence, the value of k is 3.

How many terms are in AP?

12 terms must be taken for AP.

For what value of K the terms 2k 1 7 and 3k are in AP?

Given: k, 2k – 1, 7 and 3k are in AP. Therefore, their common difference will be equal. Hence , Thus, for k = 3, given terms will be in AP.

For what value of KK 2 4k 6 3k 2 are three consecutive terms of an AP?

What is the 21st term of the AP whose first two terms are 3 and 4?

The 21st term of an AP is 137. Step-by-step explanation: It is given that first two term are -3 and 4. Therefore the 21st term of an AP is 137.

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