CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

If $$a_{1}$$, $$a_{2}$$, $$a_{3}$$ are in AP $$a_{2}$$, $$a_{3}$$, $$a_{4}$$ are in GP and $$a_{3}$$, $$a_{4}$$, $$a_{5}$$ are in HP then $$a_{1}$$, $$a_{3}$$, $$a_{5}$$ are in


A
AP
loader
B
GP
loader
C
HP
loader
D
none of these
loader

Solution

The correct option is A GP
Given $${ a }_{ 1 },{ a }_{ 2 },,{ a }_{ 3 }$$ are in A.P
Let's assume that $${ a }_{ 1 }=A-d\\ { a }_{ 2 }=A\\ { a }_{ 3 }=A+d$$
similarly $${ a }_{ 2 },{ a }_{ 3 },{ a }_{ 4 }$$ are in G.P
$$so\quad { a }_{ 2 }=A\\ { a }_{ 3 }=Ar\\ { a }_{ 4 }=A{ r }^{ 2 }$$
$$\Rightarrow A+d=Ar\\ \Rightarrow d=Ar-A$$
and finally $${ a }_{ 3 },{ a }_{ 4 },{ a }_{ 5 }$$ are in H.P
$$\Rightarrow \dfrac { 1 }{ { a }_{ 3 } } ,\dfrac { 1 }{ { a }_{ 4 } } ,\dfrac { 1 }{ { a }_{ 5 } } $$ are in A.P
hence $$\Rightarrow \dfrac { 1 }{ { a }_{ 5 } } =\dfrac { 1 }{ { a }_{ 4 } } +\left( \dfrac { 1 }{ { a }_{ 4 } } -\dfrac { 1 }{ { a }_{ 3 } }  \right) \\ \Rightarrow \dfrac { 1 }{ { a }_{ 5 } } =\dfrac { 2 }{ A{ r }^{ 2 } } -\dfrac { 1 }{ Ar } =\dfrac { 2-r }{ A{ r }^{ 2 } } \\ \Rightarrow { a }_{ 5 }=\dfrac { A{ r }^{ 2 } }{ 2-r } $$
and $${ a }_{ 1 }=A-\left( Ar-A \right) =2A-Ar=A\left( 2-r \right) \\ { a }_{ 3 }=Ar$$
clearly we can see that $${ a }_{ 1 }{ a }_{ 5 }={ { a }_{ 3 } }^{ 2 }$$
Thus $${ a }_{ 1 },{ { a }_{ 3 } },{ a }_{ 5 }$$ are in G.P

Maths

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image