If a, b, c are in AP, then 1√b+√c,1√c+√a,1√a+√b are in.
A
AP
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B
GP
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C
HP
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D
None of these
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Solution
The correct option is A AP a,b,c are in AP ⇒2b=a+c 1√c+√a−1√b+√c=1√a+√b−1√c+√a ⇒√b−√a(√c+√a)(√b+√c)=√c−√b(√a+√b)(√c+√a) cross multiplying gives ⇒b−a=c−b⇒2b=a+c ∴1√b+√c,1√c+√a,1√a+√b are in AP Hence, option A.