First slide

Arithmetico-Geometric Progression

Question

$If 10^{9} + 2 (11)^{1} (10^{8}) + 3 (11)^{2} (10)^{7} + . . + 10 (11)^{9} = k \cdot 10^{9} then k is$

Moderate

A

100
B

110
C

$\frac{121}{10}$
D

$\frac{441}{100}$

Solution

$x = 10^{9} + 2 (11) (10)^{8} + \dots + 10 (11)^{9} - - - (1)$
$\begin{array}{l} Multiplied both sides by \frac{11}{10} \\ \frac{11}{10} x = {11.10}^{8} + 2 (11)^{2} (10)^{7} + . . + 11^{10} - - - (2) \\ (1) - (2) \Rightarrow x (1 - \frac{11}{10}) = 10^{9} + 11 (10)^{8} + . . + 11^{9} - 11^{10} \\ = 10^{9} (\frac{{(\frac{11}{10})}^{10} - 1}{\frac{11}{10} - 1}) - 11^{10} (s u m o f n t e r m s i n G . P) \\ \Rightarrow - \frac{x}{10} = (11^{10} - 10^{10}) - 11^{10} = - 10^{10} \\ x = 10^{11} = 10^{9} \times 100 = k \times 10^{9} \\ \Rightarrow k = 100 \end{array}$

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