If the m^{th} terms of the series 63 + 65 + 6 | vedclass

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(C) The first series is $63, 65, 67, 69, \dots$ which is an Arithmetic Progression $(AP)$ with first term $a_1 = 63$ and common difference $d_1 = 2$.T

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$13$

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(C) The first series is $63, 65, 67, 69, \dots$ which is an Arithmetic Progression $(AP)$ with first term $a_1 = 63$ and common difference $d_1 = 2$.The $m^{th}$ term of the first series is $T_m = a_1 + (m - 1)d_1 = 63 + (m - 1)2 = 63 + 2m - 2 = 2m + 61$.The second series is $3, 10, 17, 24, \dots$ which is an $AP$ with first term $a_2 = 3$ and common difference $d_2 = 7$.The $m^{th}$ term of the second series is $T'_m = a_2 + (m - 1)d_2 = 3 + (m - 1)7 = 3 + 7m - 7 = 7m - 4$.Given that the $m^{th}$ terms are equal,we have $2m + 61 = 7m - 4$.Rearranging the terms,$61 + 4 = 7m - 2m$.$65 = 5m$.$m = 13$.

If the $m^{th}$ terms of the series $63 + 65 + 67 + 69 + \dots$ and $3 + 10 + 17 + 24 + \dots$ are equal,then $m = $

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