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

The correct option is C $$(n-1){ a }_{ 1 }{ a }_{ n }$$
Given $${ a }_{ 1 },{ a }_{ 2 },{ a }_{ 3 },......{ a }_{ n }$$ are in HP
Then, $$\cfrac { 1 }{ { a }_{ 1 } } ,\cfrac { 1 }{ { a }_{ 2 } } ,\cfrac { 1 }{ { a }_{ 3 } } ,.....\cfrac { 1 }{ { a }_{ n } } $$ will be in AP
which 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
$$\left( { a }_{ 1 }-{ a }_{ n } \right) =\cfrac { { a }_{ 1 }-{ a }_{ n } }{ { a }_{ 1 }{ a }_{ n }(n-1) } \left( { a }_{ 1 }{ a }_{ 2 }+{ a }_{ 2 }{ a }_{ 3 }+....+{ a }_{ n-1 }{ a }_{ n } \right) $$
$$\Rightarrow  \left( { a }_{ 1 }{ a }_{ 2 }+{ a }_{ 2 }{ a }_{ 3 }+....+{ a }_{ n-1 }{ a }_{ n } \right) ={ a }_{ 1 }{ a }_{ n }(n-1)\quad $$

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