Question

# Let $$a,b,c$$ be in G.P with common ratio, where $$a\ne 0$$ and $$0<r\le \cfrac { 1 }{ 2 }$$. If $$3a,7b,15c$$ are the first three terms of an A.P., then the 4th term of thisA.P is

A
73a
B
a
C
23a
D
5a

Solution

## The correct option is A $$a$$$$b=ar;c=a{ r }^{ 2 }$$$$\Rightarrow 14b=3a+15c\Rightarrow 14(ar)=3a+15a{ r }^{ 2 }\Rightarrow 14r=3+15{ r }^{ 2 }$$$$\Rightarrow 15{ r }^{ 2 }-14r+3=0\Rightarrow (3r-1)(5r-3)=0\Rightarrow r=\cfrac { 1 }{ 3 } ,\cfrac { 3 }{ 5 }$$Only acceptable value is $$r=\cfrac { 1 }{ 3 }$$, because$$r\in (0,\cfrac { 1 }{ 2 } ]\quad$$$$\therefore d=7b-3a=7r-3a=\cfrac { 7 }{ 3 } a-3a=-\cfrac { 2 }{ 3 } a\quad$$4th term $$=15c-\cfrac { 2 }{ 3 } a=\cfrac { 15 }{ 9 } a-\cfrac { 2 }{ 3 } a=a$$Maths

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