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میتونی لایو بذاری!

a) A circle is divided into 6 sectors in such a way that the angles of the sectors are in arithmetic progression. The angle of the largest sector is 4 times the angle of the smallest sector. Given that the radius of the circle is 5 cm, find the perimeter of the smallest sector.

b) The first, second and third terms of a geometric progression are $2k + 3$, $k + 6$ and $k$, respectively.

Given that all the terms of the geometric progression are positive, calculate

i) the value of the constant $k$,

ii) the sum to infinity of the progression.

پاسخ تشریحی :
نمایش پاسخ

a) $\alpha  + 5d = 4\alpha $ or $\frac{{\left( {\alpha  + 4\alpha } \right)}}{2} \times 6$

$\frac{6}{2}\left( {2\alpha  + 5d} \right)$ or $\frac{{\left( {\alpha  + 4\alpha } \right)}}{2} \times 6 = 360$

Sim Eqns $\alpha  = {24^ \circ }$ or $\frac{{2\pi }}{{15}}$ rads

Arc length $ = 5\theta $

Perimeter $ = 12.1$.

b)(i) $\frac{{k + 6}}{{2k + 3}} = \frac{k}{{k + 6}}$

$ \to {k^2} - {\text{9}}k - {\text{9}}k - {\text{36}} = 0 \to k = 12$

(NB stating $\alpha $, $\alpha r$, $\alpha {r^2}$ as $f\left( k \right)$ gets M1)

(ii) $r = {\raise0.5ex\hbox{$\scriptstyle 2$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 3$}}$,  $\alpha  = 27$

$ \to {S_\infty } = 27 \div {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 3$}} = 81$.

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