Let-a-b-c-be-the-posetive-integer-such-that-b-a-is-an-also-integer-if-a-b-c-are-in-GP-and-AM-of-a-b-c-is-b-2-then-find-the-value-of-a-2-a-14-a-1-

Question Number 17595 by virus last updated on 08/Jul/17

L e t a, b, c b e t h e p o s e t i v e i n t e g e r s u c h t h a t

b / a i s a n a l s o i n t e g e r i f a, b, c a r e i n G P a n d

A M o f a, b, c i s (b + 2) t h e n f i n d t h e v a l u e o f

(a^{2} + a - 14) / (a + 1)

Commented by virus last updated on 08/Jul/17

f i n e

Answered by mrW1 last updated on 08/Jul/17

b = na

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

\Rightarrow c = n^{2} a

\frac{a + b + c}{3} = b + 2

\Rightarrow \frac{a + na + n^{2} a}{3} = na + 2

\Rightarrow a + na + n^{2} a = 3 na + 6

\Rightarrow n^{2} - 2 n + (1 - \frac{6}{a}) = 0

n = \frac{2 + \sqrt{4 - 4 (1 - \frac{6}{a})}}{2} = 1 + \sqrt{\frac{6}{a}}

\Rightarrow a = 6

\Rightarrow n = 2 \Rightarrow b = 12 \Rightarrow c = 24

(a^{2} + a - 14) / (a + 1) = (36 + 6 - 14) / 7 = 4

Commented by virus last updated on 08/Jul/17

t h a n k y o u s i r

Commented by chux last updated on 09/Jul/17

mr W 1 \dots . please how is n = 2

Commented by mrW1 last updated on 09/Jul/17

n = 1 + \sqrt{\frac{6}{a}}

since n and a are positive integers,

the only solution is a = 6 and then

n = 2 .

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