2k - 7, k + 5 and 3k + 2 are in A.P.; the value of k is : 1. -9 2. 6 3. -7 4. 5

Given, 2k - 7, k + 5 and 3k + 2 are in A.P. ∴ (k + 5) - (2k - 7) = (3k + 2) - (k + 5) ⇒ k - 2k + 5 + 7 = 3k - k + 2 - 5 ⇒ -k + 12 = 2k - 3 ⇒ 2k + k = 12 + 3 ⇒ 3k = 15 ⇒ k = = 5. Hence, Option 4 is the correct option.

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Mathematics

2k - 7, k + 5 and 3k + 2 are in A.P.; the value of k is :
-9
6
-7
5

AP GP

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Answer

Given,

2k - 7, k + 5 and 3k + 2 are in A.P.

∴ (k + 5) - (2k - 7) = (3k + 2) - (k + 5)

⇒ k - 2k + 5 + 7 = 3k - k + 2 - 5

⇒ -k + 12 = 2k - 3

⇒ 2k + k = 12 + 3

⇒ 3k = 15

⇒ k = $\dfrac{15}{3}$ = 5.

Hence, Option 4 is the correct option.

Answered By

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Mathematics

2k - 7, k + 5 and 3k + 2 are in A.P.; the value of k is :
-9
6
-7
5

AP GP

Answer

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2k - 7, k + 5 and 3k + 2 are in A.P.; the value of k is :-96-75

AP GP

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$\dfrac{1}{a}, \dfrac{1}{b} \text{ and } \dfrac{1}{c}$ are in A.P.
Show that : bc, ca and ab are also in A.P.

If A, B and C are three arithmetic progressions (APs) as given below :
A = 2, 4, 6, 8, …….. upto n terms
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out of A + B, A - C, C - B and B - A which is/are A.P. ?
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