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1

WB JEE 2011

Subjective

The harmonic mean of two numbers is 4. Their arithmetic mean A and the geometric mean G satisfy the relation 2A + G2 = 27. Find the numbers.

Answer

.

Explanation

Let two numbers be x, y.

$$\therefore$$ $${{2xy} \over {x + y}} = 4$$, $$\sqrt {xy} = G$$ & $${{x + y} \over 2} = A$$

$$\therefore$$ $$2{G^2} = 4(2A)$$, $${G^2} = 4A$$

But $$2A + {G^2} = 27$$

$$\therefore$$ $$2A + 4A = 27$$

$$A = {{27} \over 6} \Rightarrow x + y = 9$$

$$\therefore$$ $${G^2} = 18 \Rightarrow xy = 18$$

On solving, we get x = 3, y = 6 or x = 6, y = 3.

2

WB JEE 2009

Subjective

Find the sum of the first n terms of the series 0.2 + 0.22 + 0.222 + ......

Answer

.

Explanation

Sn = 0.2 + 0.22 + 0.222 + ..... n terms

$${S_n} = {2 \over 9}(0.9 + 0.99 + 0.999 + \,\,.......\,n\,terms)$$

$${S_n} = {2 \over 9}\left( {{{10 - 1} \over {10}} + {{100 - 1} \over {100}} + {{1000 - 1} \over {1000}} + \,\,....\,n\,terms} \right)$$

$$ = {2 \over 9}\left( {1 - {1 \over {10}} + 1 - {1 \over {{{(10)}^2}}} + 1 - {1 \over {{{(10)}^3}}} + \,\,....\,n\,terms} \right)$$

$$ = {2 \over 9}\left( {n - \left( {{1 \over {10}} + {1 \over {{{(10)}^2}}} + \,\,....\,n\,terms} \right)} \right)$$

$$ = {2 \over 9}\left( {n - {{{1 \over {10}}\left( {1 - {{\left( {{1 \over {10}}} \right)}^n}} \right)} \over {1 - {1 \over {10}}}}} \right) = {2 \over 9}\left( {n - {1 \over {10}} \times {{10} \over 9}\left( {1 - {{\left( {{1 \over {10}}} \right)}^n}} \right)} \right)$$

$$ = {2 \over 9}\left( {n - {1 \over 9}\left( {1 - {{\left( {{1 \over {10}}} \right)}^n}} \right)} \right) = {{2n} \over 9} - {2 \over {81}}\left( {1 - {{\left( {{1 \over {10}}} \right)}^n}} \right)$$

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