  Question

Assertion :Suppose four distinct positive numbers $$\displaystyle a_{1}, a_{2}, a_{3}, a_{4}$$ are in GP. Let $$\displaystyle b_{1}= a_{1}, b_{2}= b_{1}+a_{2}, b_{3}=b_{2}+a_{3}$$ and $$\displaystyle b_{4}= b_{3}+a_{4}.$$STATEMENT- 1: The numbers $$\displaystyle b_{1}, b_{2}, b_{3}, b_{4}$$ are neither in AP nor in GP, and Reason: STATEMENT- 1: The numbers $$\displaystyle b_{1}, b_{2}, b_{3}, b_{4}$$ are in HP.

A
STATEMENT- 1 is True, STATEMENT- 2 is True; STATEMENT- 2 is a correct explanation for STATEMENT- 1.  B
STATEMENT- 1 is True, STATEMENT- 2 is True; STATEMENT- 2 is NOT a correct explanation for STATEMENT- 1.  C
STATEMENT- 1 is True, STATEMENT- 2 is False.  D
STATEMENT- 1 is False, STATEMENT- 2 is True.  Solution

The correct option is C STATEMENT- 1 is True, STATEMENT- 2 is False.$$\because { a }_{ 1 }, { a }_{ 2 }, { a }_{ 3 }$$ and $${ a }_{ 4}$$are in G.P with common ratio $$'r'$$$$\Rightarrow { a }_{ 1 }=a, { a }_{ 2 }=a{ r }, { a }_{ 3 }=a{ r }^{ 2 } and\quad { a }_{ 4 }=a{ r }^{ 3 }$$$$\because { b }_{ 1 }={ a }_{ 1 }=a$$$${ b }_{ 2 }={ b }_{ 1 }+{ a }_{ 2 }=a+ar=a(1+{ r })$$$${ b }_{ 3 }={ b }_{ 2 }+{ a }_{ 3 }=a(1+r)+a{ r }^{ 2 }=a(1+r+{ r }^{ 2 })$$$${ b }_{ 4 }={ b }_{ 3 }+{ a }_{ 4 }=a(1+r+{ r }^{ 2 })+a{ r }^{ 3 }=a(1+r+{ r }^{ 2 }+{ r }^{ 3 })$$$$\Rightarrow$$ Option (c) is correctMaths

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