Question

If (4k + 3 ) , ( k + 2 ), ( 3k – 7 ) and ( 2k – 3 ) are in A.P., value of k is
options:
(a) 8 / 5
(b) 8
(c)5 / 8
(d) 10

Answer 1

Answer:

Here,

⇒(4k + 3 ) , ( k + 2 ), ( 3k – 7 ) and ( 2k – 3 )

(K + 3/K + 2) = (3k - 7/2k - 3)

⇒ (K + 3) (2K – 3) = (K + 2) (3K – 7)

⇒ 2K2 – 3K + 6K – 9 = 3K2 – 7K + 6K – 14

⇒ K2 – 4K – 5 = 0

⇒ (K – 5) (K + 1) = 0

⇒ K = 5 or K = – 1

Step-by-step explanation:

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