If a 1- a 2- a 3- - a n are in H-P- then a 1 a 2- a 2 a 3- a n -1 a n will be equal toA- a1 anB- n a1 anC- n-1 a1 anD- None of these

The correct option is C Since nbsp; a_1,a_2,a_3,.................,a_n are in H.P. Therefore 1/a_1,1/a_2,1/a_3.........1/a_n will be in A.P. Which gives 1/a^ ...

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Byju's Answer

Standard XI

Mathematics

nCr Definitions and Properties

If a 1, a 2, ...

Question

If $a_{1}, a_{2}, a_{3}$ ,................., $a_{n}$ are in H.P., then $a_{1} a_{2} + a_{2} a_{3} +$ ...........+ $a_{n - 1} a_{n}$ will be equal to

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Solution

The correct option is C

Since $a_{1}, a_{2}, a_{3}$ ,................., $a_{n}$ are in H.P.

Therefore $\frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}}$ ......... $\frac{1}{a_{n}}$ will be in A.P.

Which gives $\frac{1}{a^{2}} - \frac{1}{a^{2}}$ = $\frac{1}{a^{3}} - \frac{1}{a^{2}}$ = ......... = $\frac{1}{a_{n}} - \frac{1}{a^{n - 1}}$ = d

$\Rightarrow \frac{a_{1} - a_{2}}{a_{1} a_{2}}$ = $\frac{a^{3} - a^{2}}{a_{2} a_{3}}$ =............= $\frac{a_{n - 1} - a_{n}}{a_{n - 1} a_{n}}$ = d

$\Rightarrow a_{1} - a_{2}$ = $d a_{1} a_{2}$

$a_{2} - a_{3}$ = $d a_{2} a_{3}$

....................

....................

and $a_{n - 1} - a_{n}$ = $d a_{n} a_{n - 1}$

Adding these, we get $d (a_{1} a_{2} + a_{2} a_{3} + . . . . . . . . . + a_{n})$

= ( $a_{1} + a_{2} + . . . . . . . + a_{n - 1}) - (a_{2} + a_{3} + . . . . . . + a_{n})$

= $a_{1} - a_{n}$ ..............(i)

Also $n^{t h}$ term of this A.P. is given by

$\frac{1}{a_{n}}$ = $\frac{1}{a_{1}} + (n - 1) d \Rightarrow d$ = $\frac{a_{1} - a_{n}}{a_{1} a_{n} (n - 1)}$

Substituting this value of d in (i)

$a_{1} - a_{n})$ = $\frac{a_{1} - a_{n}}{a_{1} a_{n} (n - 1)} (a_{1} a_{2} + a_{2} a_{3} + . . . . . . . + a_{n} a_{n - 1})$

$(a_{1} a_{2} + a_{2} a_{3} + . . . . . . . . . . . + a_{n} a_{n - 1})$ = $a_{1} a_{n} (n - 1)$ .

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Similar questions

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Q. If

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If a1,a2,a3,.................,an are in H.P., then a1a2+a2a3+...........+an−1an will be equal to

If $a_{1}, a_{2}, a_{3}$ ,................., $a_{n}$ are in H.P., then $a_{1} a_{2} + a_{2} a_{3} +$ ...........+ $a_{n - 1} a_{n}$ will be equal to