Sequences and Series 4 Question 5
5. If $(10)^{9}+2(11)^{1}(10)^{8}+3(11)^{2}(10)^{7}+\ldots+10(11)^{9}=k(10)^{9}$, then $k$ is equal to
(2014 Main)
(a) $\frac{121}{10}$
(b) $\frac{441}{100}$
(c) 100
(d) 110
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Solution:
- Given,
$$ \begin{aligned} & k \cdot 10^{9}=10^{9}+2(11)^{1}(10)^{8}+3(11)^{2}(10)^{7}+\ldots+10(11)^{9} \\ & \Rightarrow \quad k=1+2 \frac{11}{10}+3 \frac{11}{10}^{2}+\ldots+10 \frac{11}{10} \\ & \frac{11}{10} k=1 \quad \frac{11}{10}+2 \frac{11}{10}^{2}+\ldots+9 \frac{11}{10}^{9}+10 \frac{11}{10} \ldots \end{aligned} $$
On subtracting Eq. (ii) from Eq. (i), we get
$$ \begin{aligned} & \Rightarrow k \frac{10-11}{10}=\frac{1 \frac{11}{10}^{10}-1}{\frac{11}{10}-1}-10 \frac{11}{10}^{10} \end{aligned} $$
$\because$ In GP, sum of $n$ terms $=\frac{a\left(r^{n}-1\right)}{r-1}$, when $r>1$
$$ \begin{array}{ll} \Rightarrow & -k=1010 \frac{11}{10}^{10}-10-10 \frac{11}{10} \\ \therefore & k=100 \end{array} $$
