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$\frac{1}{{a + \frac{1}{{b + \frac{1}{c}}}}} = \frac{{16}}{{115}}$

$a,b,c \in \mathbb{N}$

Select correct option(s).

(A) a, b, c are in A.P.
(B) a, b, c are in G.P.
(C) a, b, c are in H.P.
(D) a+b+c = bc

Solution

We have, $a + \frac{1}{{b + \frac{1}{c}}} = \frac{{115}}{{16}}$

$ \Rightarrow a + \frac{c}{{bc + 1}} = \frac{{115}}{{15 + 1}}$

$ \Rightarrow a + \frac{c}{{bc + 1}} = \frac{{115}}{{5 \times 3 + 1}}$

$ \Rightarrow a + \frac{c}{{bc + 1}} = \frac{{16 \times 7 + 3}}{{5 \times 3 + 1}}$

$ \Rightarrow a + \frac{c}{{bc + 1}} = 7 + \frac{3}{{5 \times 3 + 1}}$

$\therefore \{ a,b,c\}  \equiv \{ 7,5,3\} $

Hence, (A) & (D).

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Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R $. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $ \Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$