Find the value of k.
for which the numbers 2k-1,3k+1,11 are in ap
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Answer:
Solution :
Let a = 2k+1 , b = k²+k+1 , c =3k²-3k+3
are in A.P
b - a = c - b [ common difference ]
=> (k²+k+1)-(2k+1)=(3k²-3k+3)-(k²+k+1)
=> k²+k+1-2k-1 = 3k²-3k+3-k²-k-1
=> k²-k = 2k²-4k+2
=> 0 = 2k²-4k+2-k²+k
=> k² - 3k +2 = 0
=> k² - 2k - k + 2 = 0
=> k( k - 2 ) - ( k - 2 ) = 0
=> ( k - 2 )( k - 1 ) = 0
=> k - 2 = 0 or k - 1 = 0
=> k = 2 or k = 1
•••••
Step-by-step explanation:
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