Question

If 2k+1,6,3k+1 are in a.p then find the value of ,k,

Answer 1

Solution :

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We know that ,

If a , b, c are in A.P then

b - a = c - b

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Here ,

a = 2k+1 , b = 6 , c = 3k+1 are in A.P.

b - a = c - b

=> 6 - ( 2k + 1 ) = ( 3k + 1 ) - 6

=> 6 - 2k - 1 = 3k + 1 - 6

=> -2k + 5 = 3k - 5

=> -2k - 3k = -5 - 5

=> -5k = -10

⇒ k = ( -10 )/( -5 )

⇒ K = 2

Therefore ,

k = 2

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Answer 2

HELLO DEAR,




given:- 2k + 1 , 6 , 3k + 1 are in a.p

so, we know common difference = second term - first term = third term - second term


so, 6 - (2k + 1) = (3k + 1) - 6

=> 6 - 2k - 1 = 3k + 1 - 6

=> 5 - 2k = 3k - 5

=> 5 + 5 = 3k + 2k

=> 10 = 5k

=> k = 2



I HOPE IT'S HELP YOU DEAR,
THANKS