Question

let a , b , c be positive integers such that b/a is a integer if a b c are in geometric progression and the Arithmetic mean of A,B,C is b + 2 then the value of a^2+a-14 ÷ a+1 is​

Answer 1

Answer:

$\boxed{\bold{\mathsf{ 4 }}}$

Step-by-step explanation:

According to question

To find:-

$\mathsf{ \frac{ {a}^{2} + a - 14 }{a + 1}}$

$\frac{b}{a} = \frac{c}{b} = integer$

b² = a × c

$c = \frac{ {b}^{2} }{a}$

We are given that the Airthematic mean of a,b,c is b + 2

$\implies$

$\frac{a + b + c}{3} = b + 2 \\ \\ a + b + c = 3b + 6 \\ \\ a - 2b + c = 6 \\ \\ a - 2b + \frac{ {b}^{2} }{a} = 6 \\ \\ 1 - \frac{2b}{a} + \frac{ {b}^{2} }{a} = \frac{6}{a} \\ \\ {( \frac{b}{a} - 1)}^{2} = \frac{6}{a}$

Last step we get by using (a - b)²

here , b/a is a integer

so here We will take A = 6 only because its given b/a is an integer

Now put value of a to find the answer

$\frac{ {a}^{2} + a - 14 }{a + 1} \\ \\ \frac{ {6}^{2} + 6 - 14 }{6 + 1} \\ \\ \frac{36 - 8}{7} \\ \\ = 4$

Answer 2

I hope the image will help you