If 1a- 1a-d- 1a-2d- are in HP- then the nth term of the HP is

The correct option is C a_n = 1a+(n 1)d n^th term of HP is #160; =1n^thterm of #160;the corresponding AP A.P is: a,a+d,a+2d… n^th ...

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Relation between AM,GM,HM for 2 Numbers

If 1a, 1a+d...

Question

If $\frac{1}{a}, \frac{1}{a + d}, \frac{1}{a + 2 d}, . . .$ are in HP, then the $n^{t h}$ term of the HP is

a_{n} = \frac{1}{a - (n - 1) d}

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

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

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

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$a + (a + d) + (a + 2 d) + . . . . + (a + (n - 1) d) = \frac{n}{2} [2 a + (n - 1) d]$

a_{1}, a_{2}, a_{3} . . . ., a_{n}

t a n [t a n^{- 1} (\frac{d}{1 + a_{1} a_{2}})] + t a m^{- 1} (\frac{d}{1 + a_{2} a_{3}})] + . . . + t a n^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})] =

θ = t a n^{- 1} \frac{d}{1 + a_{1} a_{2}} + t a n^{- 1} \frac{d}{1 + a_{2} a_{3}} + \dots + t a n^{- 1} \frac{d}{1 + a_{n - 1} a_{n}}

a_{1}, a_{2}, a_{3}, \dots a_{n}

t a n θ =

[a + (n - 1) d] b r^{n - 1}

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