Question

If the seventh term an AP is 1/9 Nth term is 1/7 find its 63rd term

Answer 1

$\huge\bf\green {Hey \: there!}$


$\large\color {red}\sf {\implies Given : }$


$\large\color {orange}\sf {i) The \: {7}^{th} \: term \: of \: an \: AP \: is \: { \dfrac{1}{9}}}$


$\large\color {green}\sf {ii) The \: {9}^{th} \: term \: of \: an \: AP \: is \: { \dfrac{1}{7}}}$


$\large\color {red}\sf {\implies Let :}$


$\large\color {orange}\sf {i) First \: term = a}$


$\large\color {green}\sf {ii) Common \: difference = d}$


$\large\color{purple}\sf {\implies According \: to \: the \: Question :}$


$\large\color {blue}\sf {\implies a_7 = a_1 + ( 7 - 1 )d = {\dfrac {1}{9}}}$


$\large\color {blue}\sf {\implies a_7 = a_1 + 6d = {\dfrac {1}{9} \rightarrow \: 1}}$


$\large\color {darkred}\sf {\implies a_9 = a_1 + ( 9 - 1 )d={\dfrac {1}{7}}}$


$\large\color {darkred}\sf {\implies a_9 = a_1 + 8d ={\dfrac {1}{7} \rightarrow \: 2}}$


$\large\color {violet}\sf {Subtracting \: eq \: ( 2 ) \: from \: ( 1 ) , we \: have :}$


$\large\color {blue}\sf {a_1 + 6d = {\dfrac {1}{9}} }$


$\large\color {darkred}\sf {a_1 + 8d = {\dfrac {1}{7}}}$

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$\large\color {green}\sf {- 2d = {\dfrac {1}{9}} - {\dfrac {1}{7}}}$


$\large\color {green} \sf{ \implies - 2d = {\dfrac {- 2}{63}}}$


$\large\color {green}\sf {\implies d ={ \dfrac {1}{63}}}$


$\large\color {violet}\sf {Substituting \: value \: of \: d \: in \: eq ( 1 ), we \: get : }$


$\large\color {yellow}\sf { \implies a_1 + 6 ({\dfrac {1}{63}}) = {\dfrac {1}{9}}}$


$\large\color {yellow}\sf {\implies a_1 = {\dfrac {1}{9} - {\dfrac {6}{63}}}}$


$\large\color {yellow}\sf {\implies a_1 = {\dfrac{1}{63}}}$


$\large\color {green}\sf {Now,}$


$\large\color {orange}\sf {\implies \: a(63)= {\dfrac {1}{63}} + (63 - 1) × {\dfrac {1}{63}}}$


$\large\color {orange}\sf {\implies a (63) = {\dfrac {1}{63}} + {\dfrac {62}{63}}}$


$\large\color {orange}\sf {\implies a (63) = 1}$


$\large\color {violet}\sf {\implies Formula \: used :}$


$\large\color {purple}\sf {1. \: T_n = a_1 + ( a_n - 1 )d}$


$\mathbb {\huge {\fcolorbox {yellow}{green}{Hope It Helps }}}$


$\huge\bf\red {\#BeBrainly}$

Answer 2

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\bf we \: get$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\bf \implies T_{63} = 1$

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