how to solve this tell me step by step
Is series √3,√6,√9,√12..... an AP? Give reason
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Answered by
0
1√3, 2√3,3,3√4......
now subtract any two
2√3-1√3
√3
on solving further...u will not get same difference...so it is not in AP
now subtract any two
2√3-1√3
√3
on solving further...u will not get same difference...so it is not in AP
ReetChauhan1112:
2√3 = √12
so this is wrong
× wrong
Answered by
1
√3,√6,√9,√12....
= √3,(√2√3),(√3√3),(√4√3)
let a = √3
b = √2√3
c = √3√3
d = √4√3
so,
D1 = b-a =√2√3-√3 = √3(√2-1)
D2 = c-b = √3√3-√2√3 = √3(√3-√2)
here,
D1≠D2
so, these are not in AP
= √3,(√2√3),(√3√3),(√4√3)
let a = √3
b = √2√3
c = √3√3
d = √4√3
so,
D1 = b-a =√2√3-√3 = √3(√2-1)
D2 = c-b = √3√3-√2√3 = √3(√3-√2)
here,
D1≠D2
so, these are not in AP
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