Find the 9^{th} term of the series frac{1}{7} | vedclass

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(A) The given series is $\frac{1}{7}, \frac{1}{14}, \frac{1}{21}, \frac{1}{28}, \dots$ This is a harmonic progression because the reciprocals of the t

Vedclass · Accepted Answer

$1/63$

Vedclass · Answer

(A) The given series is $\frac{1}{7}, \frac{1}{14}, \frac{1}{21}, \frac{1}{28}, \dots$ This is a harmonic progression because the reciprocals of the terms form an arithmetic progression: $7, 14, 21, 28, \dots$ For the arithmetic progression,the first term $a = 7$ and the common difference $d = 14 - 7 = 7$. The $n^{th}$ term of an arithmetic progression is given by $a_n = a + (n - 1)d$. For $n = 9$,$a_9 = 7 + (9 - 1) 	imes 7 = 7 + 8 	imes 7 = 7 + 56 = 63$. Therefore,the $9^{th}$ term of the harmonic progression is the reciprocal of $a_9$,which is $\frac{1}{63}$.

Find the $9^{th}$ term of the series $\frac{1}{7} + \frac{1}{14} + \frac{1}{21} + \frac{1}{28} + \dots$

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