Question

# If $${ a }_{ 1 },{ a }_{ 2 },{ a }_{ 3 },....{ a }_{ n }$$ are in HP, then $${ a }_{ 1 }{ a }_{ 2 }+{ a }_{ 2 }{ a }_{ 3 }+.....+{ a }_{ n-1 }{ a }_{ n }$$ will be equal to

A
a1an
B
na1an
C
(n1)a1an
D
None of these

Solution

## The correct option is C $$(n-1){ a }_{ 1 }{ a }_{ n }$$Given $${ a }_{ 1 },{ a }_{ 2 },{ a }_{ 3 },......{ a }_{ n }$$ are in HPThen, $$\cfrac { 1 }{ { a }_{ 1 } } ,\cfrac { 1 }{ { a }_{ 2 } } ,\cfrac { 1 }{ { a }_{ 3 } } ,.....\cfrac { 1 }{ { a }_{ n } }$$ will be in APwhich gives$$\cfrac { 1 }{ { a }_{ 2 } } -\cfrac { 1 }{ { a }_{ 1 } } =\cfrac { 1 }{ { a }_{ 3 } } -\cfrac { 1 }{ { a }_{ 2 } } =.....=\cfrac { 1 }{ { a }_{ n } } -\cfrac { 1 }{ { a }_{ n-1 } } =d$$$$\Rightarrow \cfrac { { a }_{ 1 }-{ a }_{ 2 } }{ { a }_{ 1 }{ a }_{ 2 } } =\cfrac { { a }_{ 2 }-{ a }_{ 3 } }{ { a }_{ 2 }{ a }_{ 3 } } =......=\cfrac { { a }_{ n-1 }-{ a }_{ n } }{ { a }_{ n-1 }{ a }_{ n } } =d$$$$\Rightarrow { a }_{ 1 }-{ a }_{ 2 }=d{ a }_{ 1 }{ a }_{ 2 }$$$${ a }_{ 2 }-{ a }_{ 3 }=d{ a }_{ 2 }{ a }_{ 3 }$$............$${ a }_{ n-1 }-{ a }_{ n }=d{ a }_{ n-1 }{ a }_{ n }$$On adding these, we have$$d\left( { a }_{ 1 }{ a }_{ 2 }+{ a }_{ 2 }{ a }_{ 3 }+....+{ a }_{ n-1 }{ a }_{ n } \right) ={ a }_{ 1 }-{ a }_{ n }$$     ....(i)Also $$n^{th}$$ term of this AP is given by$$\cfrac { 1 }{ { a }_{ n } } =\cfrac { 1 }{ { a }_{ 1 } } +(n-1)d\quad$$$$d=\cfrac { { a }_{ 1 }-{ a }_{ n } }{ { a }_{ 1 }{ a }_{ n }(n-1) }$$On substituting this value of $$d$$ in Eq. (i), we get$$\left( { a }_{ 1 }-{ a }_{ n } \right) =\cfrac { { a }_{ 1 }-{ a }_{ n } }{ { a }_{ 1 }{ a }_{ n }(n-1) } \left( { a }_{ 1 }{ a }_{ 2 }+{ a }_{ 2 }{ a }_{ 3 }+....+{ a }_{ n-1 }{ a }_{ n } \right)$$$$\Rightarrow \left( { a }_{ 1 }{ a }_{ 2 }+{ a }_{ 2 }{ a }_{ 3 }+....+{ a }_{ n-1 }{ a }_{ n } \right) ={ a }_{ 1 }{ a }_{ n }(n-1)\quad$$Maths

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