if 2k+1,6,3k+1 is an A.P then find the value of k
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Answered by
8
if 2k+1,6,3k+1 are in AP then
(2k+1)+(3k+1) = 2×6
[a+c = 2b]
5k+2 = 12
5k = 10
k = 2
(2k+1)+(3k+1) = 2×6
[a+c = 2b]
5k+2 = 12
5k = 10
k = 2
Answered by
6
since the given numbers are in AP
The common difference is constant
T3-T2 = T2-T1
=> 3k +1-6 = 6 -2k -1
=> 5k -5 = 5
=> 5k =10
=> k =2
The common difference is constant
T3-T2 = T2-T1
=> 3k +1-6 = 6 -2k -1
=> 5k -5 = 5
=> 5k =10
=> k =2
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