If a2, b2, c2, are in AP, then show that a/b+c, b/c+a, c/a+b are in AP - Maths - Sequences and Series

Question

If a2, b2, c2, are in AP, then show that a/b+c, b/c+a, c/a+b are
in AP

Zico · Accepted Answer

Given that $n n n^{n a n n n 2 n n} n, n n^{n b n n n 2 n n} n, n n^{n c n n n 2 n n} n n$ are in A.P.

We need to prove that $n n \frac{n}{n a n} n, n \frac{n}{n b n} n, n \frac{n}{n c n} n n$ are in A.P.

Since $n n n^{n a n n n 2 n n} n, n n^{n b n n n 2 n n} n, n n^{n c n n n 2 n n} n n$ are in A.P.

$n n n^{n 2 n b n n n 2 n n} n = n n^{n a n n n 2 n n} n + n n^{n c n n n 2 n n} n n n^{n b n n n 2 n n} n - n n^{n a n n n 2 n n} n = n n^{n c n n n 2 n n} n - n n^{n b n n n 2 n n} n n n (n b n + n a n) n n (n b n - n a n) n n n = n (n c n + n b n) n n (n c n - n b n) n n n n n \frac{n}{n (n b n - n a n) n} n n n = n n \frac{n}{n (n c n - n b n) n} n n n n n \frac{n}{n (n b n - n a n) n} n n n = n n \frac{n}{n (n c n - n b n) n} n n n n n \frac{n}{n b n + n c n - n c n - n a n} n n n = n n \frac{n}{n (n c n + n a n - n a n - n b n) n} n n n \frac{n}{n 1 n} n - n \frac{n}{n 1 n} n = n \frac{n}{n 1 n} n - n \frac{n}{n 1 n} n n$

That is,

$n n \frac{n}{n 1 n} n, n \frac{n}{n 1 n} n, n \frac{n}{n 1 n} n n$ are in A.P.

Multiply each term by $n n n (n a n + n b n + n c n) n n n$ .

Thus,

$n n \frac{n}{n n (n a n + n b n + n c n) n n} n, n \frac{n}{n n (n a n + n b n + n c n) n n} n, n \frac{n}{n n (n a n + n b n + n c n) n n} n n$ are in A.P.

That is,

$n n \frac{n}{n a n} n + n 1 n, n \frac{n}{n b n} n + n 1 n, n \frac{n}{n c n} n + n 1 n n$ are in A.P.

That is

$n n \frac{n}{n a n} n, n \frac{n}{n b n} n, n \frac{n}{n c n} n n$ are in A.P

If a2, b2, c2, are in AP, then show that a/b+c, b/c+a, c/a+b are in AP.