Question

if a^2 b^2 c^2 are in ap then a/b+c b/c+a c/a+b are in ?

Answer 1

Answer:

please answer questions no. 16 to 19 please

Answer 2

Answer:

AP

Step-by-step explanation:

∵ a sequence is called AP if there is a common difference in consecutive terms,

If a², b² ,c² are in AP,

$\implies b^2-a^2=c^2-b^2$

$(b+a)(b-a)=(c+b)(c-b)-----(1)$

$\frac{b}{c+a}-\frac{a}{b+c}=\frac{b^2+bc-ac-a^2}{(c+a)(b+c)}$

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

$=\frac{(b+a)(b-a)+c(b-a)}{(c+a)(b+c)}$

$=\frac{(b-a)(a+b+c)}{(c+a)(b+c)}$

From equation (1),

$=\frac{\frac{(c+b)(c-b)}{(b+a)}(a+b+c)}{(c+a)(b+c)}$

$=\frac{(c-b)(a+b+c)}{(c+a)(b+a)}$

Now,

$\frac{c}{a+b}-\frac{b}{c+a}$

$=\frac{c(c+a)-b(a+b)}{(a+b)(c+a)}$

$=\frac{c^2+ac-ab-b^2}{(a+b)(c+a)}$

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

$=\frac{(c-b)(a+b+c)}{(a+b)(c+a)}$

Hence,

$\frac{b}{c+a}-\frac{a}{b+c}=\frac{c}{a+b}-\frac{b}{c+a}$

Therefore,

$\frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}\text{ are in AP}$