If $\frac{1}{(a + 1)}, \frac{1}{3 a}, \frac{1}{(4 a + 2)}$ are in harmonic progression, then the value of $a$ will be

\frac{1}{5}

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\frac{1}{2}

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\frac{1}{3}

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\frac{1}{4}

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Solution

The correct option is C $\frac{1}{3}$
The reciprocal of A.P. is H.P.
A..P is in the form of $a + 1, 3 a, 4 a + 2, . . .$
Common difference, $d =$ second term $-$ first term
$3 a - (a + 1) = 4 a + 2 - 3 a$
$\Rightarrow 2 a - 1 = a + 2$
$\Rightarrow a = 3$
Therefore, the value of $a$ in H..P is $\frac{1}{3}$ .

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Similar questions

(a + 1), 3 a, (4 a + 2)

a

a = x + 1

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x

z_{1}, z_{2}, z_{3},

z_{1}, z_{2}, z_{3},

\frac{1}{a}

\frac{1}{b}

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\frac{1}{d}

a_{1}, a_{2}, a_{3} \dots, a_{9}

a_{4} = 5

a_{5} = 4,

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 / a_{1} & 1 / a_{2} & 1 / a_{3} 1 / a_{4} & 1 / a_{5} & 1 / a_{6} 1 / a_{7} & 1 / a_{8} & 1 / a_{9} \end{matrix} ∣ ∣ ∣ ∣ ∣

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