SATHEE: Sequences and Series 4 Question 5

Sequences and Series 4 Question 5

5.

If $(10)^{9} + 2 (11)^{1} (10)^{8} + 3 (11)^{2} (10)^{7} + \dots + 10 (11)^{9} = k (10)^{9}$ , then $k$ is equal to

(2014 Main)

(a) $\frac{121}{10}$

(b) $\frac{441}{100}$

(d) 110

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Answer:

Correct Answer: 5. (c)

Solution:

Given,

$\begin{aligned} k \cdot 10^{9} = 10^{9} + 2 (11)^{1} (10)^{8} + 3 (11)^{2} (10)^{7} + \dots + 10 (11)^{9} \\ \Rightarrow k = 1 + 2 \frac{11}{10} + 3 {\frac{11}{10}}^{2} + \dots + 10 \frac{11}{10} \\ \frac{11}{10} k = 1 \frac{11}{10} + 2 {\frac{11}{10}}^{2} + \dots + 9 {\frac{11}{10}}^{9} + 10 \frac{11}{10} \dots \end{aligned}$

On subtracting Eq. (ii) from Eq. (i), we get

$\begin{aligned} \Rightarrow k \frac{10 - 11}{10} = \frac{1 {\frac{11}{10}}^{10} - 1}{\frac{11}{10} - 1} - 10 {\frac{11}{10}}^{10} \end{aligned}$

$[∵$ In GP, sum of $n$ terms $= \frac{a (r^{n} - 1)}{r - 1}$ , when $r > 1]$

$\begin{array}{ll} \Rightarrow & - k = 10 [10 (\frac{11}{10})^{10} - 10 - 10 (\frac{11}{10})^{10}] \\ ∴ & k = 100 \end{array}$

Sequences and Series 4 Question 5

5.

Answer:

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Sequences and Series 4 Question 5

5.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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