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Question

Suppose four distinct positive numbers $$\mathrm{a}_{1},\ \mathrm{a}_{2},\ \mathrm{a}_{3},\ \mathrm{a}_{4}$$ are in $$G.\mathrm{P}$$. Let $$\mathrm{b}_{1}=\mathrm{a}_{1},\ \mathrm{b}_{2}=\mathrm{b}_{1}+\mathrm{a}_{2},\ \mathrm{b}_{3}=\mathrm{b}_{2}+\mathrm{a}_{3}$$ and $$\mathrm{b}_{4}=\mathrm{b}_{3}+\mathrm{a}_{4}$$.
STATEMENT-1 : The numbers $$\mathrm{b}_{1},\ \mathrm{b}_{2},\ \mathrm{b}_{3},\ \mathrm{b}_{4}$$ are neither in $$A.\mathrm{P}$$. nor in $$G.\mathrm{P}$$.
STATEMENT-2 : The numbers $$\mathrm{b}_{1},\ \mathrm{b}_{2},\ \mathrm{b}_{3},\ \mathrm{b}_{4}$$ are in $$\mathrm{H}.\mathrm{P}$$.


A
STATEMENT1 is True, STATEMENT2 is True; STATEMENT2 is a correct explanation for STATEMENT1
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B
STATEMENT1 is True, STATEMENT2 is True; STATEMENT2 is NOT a correct explanation for STATEMENT1.
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C
STATEMENT1 is True, STATEMENT2 is False
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D
STATEMENT1 is False, STATEMENT2 is True
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Solution

The correct option is B STATEMENT1 is True, STATEMENT2 is False
$$a1=k, a2=kr, a3=k{ r }^{ 2 }, a4=k{ r }^{ 3 }$$
$$b1=k, b2=k+kr, b3=k+kr+k{ r }^{ 2 }, b4=k+kr+k{ r }^{ 2 }+k{ r }^{ 3 }$$
Thus, it is obvious that $$b1, b2, b3, b4$$ are neither in AP nor in GP.
Again, $$\dfrac { 1 }{ b1 } +\dfrac { 1 }{ b3 } =\dfrac { 1 }{ k } +\dfrac { 1 }{ k+kr+k{ r }^{ 2 } } \neq \dfrac { 2 }{ k+kr } =\dfrac { 2 }{ b2 } $$
Hence, $$b1, b2, b3$$ are not in HP.
Hence, (C) is correct.

Mathematics

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