if 2k+1,6,3k+1 are in terms of ap then find the value of k
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NotDNerdType; Ambitious. The given terms are in A.P. Hence, (6)-(2k+1) = ( 3k+1)-(6) = d -2k+5 = 3k-5 10 = 5k k= 2. Thankyou!!
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6-2k+1=6-3k+1
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a₁=2k+1
a₂=6
a₃=3k+1
Difference is equal throughout an A.P
So,a₂-a₁=a₃-a₂
6-2k-1=3k+1-6
-2k+5=3k-6
3k+2k=5+5
5k=10
k=10/5
k=2
The value of k is 2.
a₂=6
a₃=3k+1
Difference is equal throughout an A.P
So,a₂-a₁=a₃-a₂
6-2k-1=3k+1-6
-2k+5=3k-6
3k+2k=5+5
5k=10
k=10/5
k=2
The value of k is 2.
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