Question

For what value of k, 2k-7 , k+5 and 3k+2 are three consecutive terms of an A.P ?

Answer 1

The value of k is 5.

Common difference: d= 7

Terms are 3,10 and 17.

Given:

• Three terms.
• 2k-7 , k+5 and 3k+2

To find:

• For what value of k, are the given three consecutive terms of an A.P ?

Solution:

Concept to be used:

1. Common difference (d) is difference of consecutive terms.
2. Common difference is same if consecutive terms of an A.P.

Step 1:

Find the common difference.

Let

$a = 2k - 7$

$a_1 = k + 5$

or

$a_2 = 3k + 2$

Thus,

$d = a_1 - a = k + 5 - 2k + 7$

or

$\bf d= - k + 12 \: \: ...eq1$

or

$a_2 - a_1 = d= 3k + 2 - k - 5$

or

$\bf d= 2k - 3 \: \: ...eq2$

Step 2:

Equate both equations.

$- k + 12 = 2k - 3$

or

$- k - 2k = - 3 - 12$

or

$3k = 15$

or

$\bf k = 5$

Thus,

Value of k is 5 for the terms to be conservative terms of A.P.

Common difference: d= 7

Terms are 3,10,17.

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Answer 2

Given,

Three consecutive terms of an A.P.;

2k-7, k+5 and 3k+2

To find,

The value of k.

Solution,

We can simply solve this mathematical problem using the following process:

The three consecutive terms of the given A.P.;

The first term of the A.P. = 2k-7

The second term of the A.P. = k+5

The third term of the A.P. = 3k+2

Mathematically,

For three consecutive terms of an A.P.,

For three consecutive terms of an A.P., the second term in the A.P. is equal to the average of the first and third terms.

Now, in the given A.P.;

The second term = average of the first and third term

=> (k+5) = {(2k-7) + (3k+2)}/2

=> {2k-7 + 3k+2} = 2(k+5)

=> 5k - 5 = 2k + 10

=> 3k = 15

=> k = 5

Hence, the value of k is equal to 5.