If a 1- a 2- a 3- - a 10 are in H-P- then the value of 1- a 1 a 10 a 1 a 2- a 2 a 3- a 3 a 4- a 9 a 10 is

A_1, a_2, a_3, … are H.P. 1a_1, 1a_2, 1a_3, … are A.P. Let the common difference be d, nbsp; Now d= 1a_2 1a_1 =1a_3 1a_2= … = 1a_10 1a_9 nbsp; So, a_1 ...

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Standard XI

Mathematics

Nth term of HP

If a 1, a 2, ...

Question

If $a_{1}, a_{2}, a_{3}, \dots a_{10}$ are in H.P., then the value of $\frac{1}{a_{1} a_{10}} (a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} + \dots + a_{9} a_{10})$ is

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Solution

$a_{1}, a_{2}, a_{3}, \dots$ are H.P.
$\frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}}, \dots$ are A.P.
Let the common difference be $d$ ,
Now
$d = \frac{1}{a_{2}} - \frac{1}{a_{1}} = \frac{1}{a_{3}} - \frac{1}{a_{2}} = \dots = \frac{1}{a_{10}} - \frac{1}{a_{9}}$
So,
$a_{1} a_{2} = \frac{a_{1} - a_{2}}{d} a_{2} a_{3} = \frac{a_{2} - a_{3}}{d} . . . a_{9} a_{10} = \frac{a_{9} - a_{10}}{d}$
Now,
$\frac{1}{a_{1} a_{10}} (a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} + \dots + a_{9} a_{10}) = \frac{1}{a_{1} a_{10}} (\frac{a_{1} - a_{2}}{d} + \frac{a_{2} - a_{3}}{d} + \dots + \frac{a_{9} - a_{10}}{d}) = \frac{a_{1} - a_{10}}{(a_{1} a_{10}) d} = \frac{1}{d} (\frac{1}{a_{10}} - \frac{1}{a_{1}}) = \frac{1}{d} (\frac{1}{a_{1}} + 9 d - \frac{1}{a_{1}}) = 9$

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Q. If

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Nth term of HP

Standard XI Mathematics

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If a1,a2,a3,…a10 are in H.P., then the value of 1a1a10(a1a2+a2a3+a3a4+⋯+a9a10) is

a1,a2,a3,… are H.P. 1a1,1a2,1a3,… are A.P. Let the common difference be d, Now d=1a2−1a1=1a3−1a2=…=1a10−1a9 So, a1a2=a1−a2da2a3=a2−a3d...a9a10=a9−a10d Now, 1a1a10(a1a2+a2a3+a3a4+⋯+a9a10)=1a1a10(a1−a2d+a2−a3d+⋯+a9−a10d)=a1−a10(a1a10)d=1d(1a10−1a1)=1d(1a1+9d−1a1)=9

If $a_{1}, a_{2}, a_{3}, \dots a_{10}$ are in H.P., then the value of $\frac{1}{a_{1} a_{10}} (a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} + \dots + a_{9} a_{10})$ is