if 2k ,3k-2,3k+2 are three consecutive terms of an AP then find the value of k
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if a,b, c are in A.P. then
2b = a + c
now according to question
2(3k - 2) = (2k) + (3k+2)
6k - 4 = 2k + 3k + 2
6k - 4 = 5k + 2
6k - 5k = 2 + 4
k = 6 ...Answer
Hope it will help you.✌✌
2b = a + c
now according to question
2(3k - 2) = (2k) + (3k+2)
6k - 4 = 2k + 3k + 2
6k - 4 = 5k + 2
6k - 5k = 2 + 4
k = 6 ...Answer
Hope it will help you.✌✌
aaryan27:
when its not AP how u can say its an AP
if we calculate the difference i.e a2-a1 and a3-a2 d1 is not equal to d2
okay thanks bro
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