Math, asked by BSuday, 1 year ago

if t7 is equal to 1 by 9 and t9 is 1 by 7 find the 63rd term in AP

Answers

Answered by samrat00725100
6
t_7 = \frac{1}{9} \text{ and } t_9 = \frac{1}{7}
We can write t_7 = t_1 + 6d and t_9 = t_1 + 8d
\therefore t_1 + 6d = \frac{1}{9} \text{ and } t_1 + 8d = \frac{1}{7}
Solving this two equations we get :-
 t_1 = \frac{1}{63} \text{ and } d = \frac{1}{63}

Thus if we need to find the 63^{rd} term in AP...

 t_{63} = t_1 + 62d
 t_{63} = \frac{1}{63} + 62 \cdot \frac{1}{63}
 t_{63} = \frac{1}{63} (1 + 62)
 t_{63} = \frac{1}{\cancel{63}} \cdot \cancel{63}
 t_{63} = 1
Answered by Anonymous
4

 \huge \bf Solution :

Let a be the first term of and d be the common difference of the given AP. Then,

 \bf T_{7} = \dfrac{1}{9}

 \bf \implies a + 6d = \dfrac{1}{9} - ①

 \bf T_{9} = \dfrac{1}{7}

 \bf \implies a + 8d = \dfrac{1}{7} - ②

 \bf On \: subtracting \: ① \: from \: ②, \: we \: get

 \bf  (a + 8d) - (a + 6d) =  \dfrac{1}{7} - \dfrac{1}{9}

 \bf a + 8d - a - 6d = \dfrac{9}{63} - \dfrac{7}{63}

 \bf 2d = \dfrac{9 - 7}{63}

 \bf 2d = \dfrac{2}{63}

 \bf d = \dfrac{2}{63 \times 2}

 \bf d = \dfrac{\cancel{2}}{63 \times \cancel{2}}

 \bf d = \dfrac{1}{63}

 \large \boxed{\bf d = \dfrac{1}{63}}

 \bf By, \: putting \: d = \dfrac{1}{63} \: in \: ①,

 \bf we \: get

 \bf a + 6 \times \dfrac{1}{63} = \dfrac{1}{9}

 \bf a + \cancel{6}^{2} \times \dfrac{1}{\cancel{63}_{21}} = \dfrac{1}{9}

 \bf a + \dfrac{2}{21} = \dfrac{1}{9}

 \bf a = \dfrac{1}{9} - \dfrac{2}{21}

 \bf a = \dfrac{7}{63} - \dfrac{6}{63}

 \bf a = \dfrac{7 - 6}{63}

 \bf a = \dfrac{1}{63}

 \large \boxed{\bf a = \dfrac{1}{63}}

 \bf Thus, \: a = \dfrac{1}{63} \: and \: d = \dfrac{1}{63}

 \bf \therefore T_{63} = a + (63 - 1)d = a + 62d

 \bf \implies T_{63} = \dfrac{1}{63} + 62 \times \dfrac{1}{63}

 \bf \implies T_{63} = \dfrac{1}{63} + \dfrac{62}{63}

 \bf \implies T_{63} = \dfrac{1 + 62}{63}

 \bf \implies T_{63} = \dfrac{63}{63}

 \bf \implies T_{63} = \cancel{\dfrac{63}{63}}

 \bf \implies T_{63} = 1

 \large \boxed{\bf T_{63} = 1 }

 \bf Hence, \: 63rd \: term \: of \: given \: AP \: is \: 1.

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