6 If a2, b2,c2 are in AP then show that ab+c, bc+a, ca+b8 If b+c-aa, c+a-bb, a+b-cc - Maths - Arithmetic Progressions

Question

6   I f   a 2 ,   b 2 , c 2   a r e   i n
  A P   t h e n   s h o w   t h a t   a b
+ c ,   b c + a ,   c a + b 8   I f   b + c - a
a ,   c + a - b b ,   a + b - c c   a r e   i n
  A P   t h e n   s h o w   t h a t   1 a
,   1 b ,   1 c   a r e   i n   A P  
i f   a + b + c ≠ 0

Brijendra Pal · Accepted Answer

Hi,
6th:

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

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6 . I f a 2 , b 2 , c 2 a r e i n A P . t h e n s h o w t h a t a b + c , b c + a , c a + b 8 . I f b + c - a a , c + a - b b , a + b - c c a r e i n A P t h e n s h o w t h a t 1 a , 1 b , 1 c a r e i n A P i f a + b + c ≠ 0