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Let $\left\{a_k\right\}$ and $\left\{b_k\right\}, k \in N$, be two G.P.s with common ratio $r_1$ and $r_2$ respectively such that $a_1=b_1=4$ and $r_1 < r_2$. Let $c_k=a_k+k, \in N$. If $c_2=5$ and $c_3=13 / 4$ then $\sum \limits_{k=1}^{\infty} c_k - \left(12 a _6+8 b _4\right)$ is equal to
$9$
$18$
$20$
$22$
Solution
Given that
$c_k=a_k+b_k$ and
also
$a _2=4 r _1$ $\quad a _3=4 r _1{ }^2$
$b _2=4 r _2$ $\quad b _3=4 r _2{ }^2$
Now $c_2=a_2+b_2=5$ and $c_3=a_3+b_3=\frac{13}{4}$
$\Rightarrow r_1+r_2=\frac{5}{4}$ and $r_1^2+r_2^2=\frac{13}{16}$
Hence $r_1 r_2=\frac{3}{8}$ which gives $r_1=\frac{1}{2} \quad \& \quad r_2=\frac{3}{4}$
$\sum \limits_{ k =1}^{\infty} c _{ k }-\left(12 a _6+8 b _4\right)$
$=\frac{4}{1-r_1}+\frac{4}{1-r_2}-\left(\frac{48}{32}+\frac{27}{2}\right)$
$=24-15=9$
