for what value of k will the consecutive terms 2k+1,3k+3,5k-3 form an AP?
Answers
Answered by
3
any series in AP when common difference of this series is always constant .
now ,
=>3k+3-2k-1=5k-3-3k-3
=>k+2 =2k-6
=> k=8
now ,
=>3k+3-2k-1=5k-3-3k-3
=>k+2 =2k-6
=> k=8
aqibshaikh:
2k = 8
its a mistake in ur answer
and in the first step -3 will be +3 due to minus sign
well its ok
u r righy
Answered by
1
here, 2k+1 , 3k +3, and 5k-3 forms an AP.
k = 8
k = 8
please again check your answer I'm right
u open bracket with -3 same in the question but my dear u r expert u should now - * - will be +3
Um actually I think ur answer is wrong cuz 3k+3-(2k+1) = 3k+3-2k-1
hm yes got my mistake. thanks
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