Question

Find the value of n if
$5 + 9 + 13 + ...........to \: n \: terms \: \div 7 + 9 + 11 + .............to \: {n + 1} \: terms = 17 \div 16|$

Answer 1

We know that the sum of 'n' terms of an A.P is given by :

✿  $\bf{S_n = \frac{n}{2}[2a + (n - 1)d]}$

Let us consider the Series in the Numerator :

✿  5 + 9 + 13 + . . .to n terms

In Numerator series :

✡  First term (a) = 5

✡  Common Difference (d) : Second term - First term = (9 - 5) = 4

$\bf{Sum\;of\;n\;terms\;of\;Numerator\;series = \frac{n}{2}[2(5) + (n - 1)4]}$

$\bf{Sum\;of\;n\;terms\;of\;Numerator\;series = \frac{n}{2}[10 + (n - 1)4]}$

$\bf{Sum\;of\;n\;terms\;of\;Numerator\;series = \frac{n}{2}(2)[5 + (n - 1)2]}$

$\bf{Sum\;of\;n\;terms\;of\;Numerator\;series = n[5 + 2n - 2]}$

$\bf{Sum\;of\;n\;terms\;of\;Numerator\;series = n[2n + 3]}$

$\bf{Sum\;of\;n\;terms\;of\;Numerator\;series = 2n^2 + 3n}$

Let us consider the Series in the Denominator :

✿  7 + 9 + 11 + . . .to (n + 1) terms

In Denominator series :

✡  First term = 7

✡  Common Difference (d) = (9 - 7) = 2

$\bf{Sum\;of\;(n + 1)\;terms\;of\;Denominator\;series = (\frac{n + 1}{2})[2(7) + (n + 1 - 1)2]}$

$\bf{Sum\;of\;(n + 1)\;terms\;of\;Denominator\;series = (\frac{n + 1}{2})[2(7) + 2n]}$

$\bf{Sum\;of\;(n + 1)\;terms\;of\;Denominator\;series = (\frac{n + 1}{2})(2)[7 + n]}$

$\bf{Sum\;of\;(n + 1)\;terms\;of\;Denominator\;series = (n + 1)(n + 7)}$

$\bf{Sum\;of\;(n + 1)\;terms\;of\;Denominator\;series = n^2 + 8n + 7}$

The Question reduces to :

$\bf{\implies \frac{2n^2 + 3n}{n^2 + 8n + 7} = \frac{17}{16}}$

$\bf{\implies 16(2n^2 + 3n) = 17(n^2 + 8n + 7)}$

$\bf{\implies 32n^2 + 48n = 17n^2 + 136n + 119}$

$\bf{\implies 15n^2 - 88n - 119 = 0}$

$\bf{\implies 15n^2 - 105n + 17n - 119 = 0}$

$\bf{\implies 15n(n - 7) + 17(n - 7) = 0}$

$\bf{\implies (n - 7)(15n + 17) = 0}$

$\bf{\implies n = 7\;(or)\;n = \frac{-17}{15}}$

$\bf{As\;n\;cannot\;be\;(\frac{-17}{15})\;[because\;n\;is\;positive\;number]}$

$\bf{\implies n = 7}$

Answer 2

$\underline{\underline{\mathfrak{\Large{Solution : }}}}$




$\underline{\textsf{Given, }} \\ \\ \sf \implies \dfrac{5 \: + \: 9 \: + \: 13 \: + ..... n \: terms}{7 \: + \: 9 \: + \: 11 \: + ..... ( n \: + \: 1 ) terms }\: = \: \dfrac{17}{16}$





$\textsf{Here, the numerator as well as denominator} \\ \textsf{forms A.P.}$




$\underline{\textsf{For Numerator : }} \\ \\ \sf \implies First \: term (a_1) \: = \: 5 \\ \\ \sf \implies Common \: difference (d_1) \: = \: 9 \: - \: 5 \: = \: 4 \\ \\ \sf \implies No. \: of \: terms \: = \: n \\ \\ \\ \underline{\textsf{For Denominator :} } \\ \\ \sf \implies First \: term (a_2) \: = \: 7 \\ \\ \sf \implies Common \: difference (d_2) \: = \: 9 \: - \: 7 \: = \: 2 \\ \\ \sf \implies No. \: of \: terms \: = \: n \: + \: 1$






$\underline{\textsf{Using Formula : }} \\ \\ \sf \implies S_n \: = \: \dfrac{n}{2}\{ 2a \: + \: ( n \: - \: 1 )d \}$



$\sf \implies \dfrac{ \quad \dfrac{n}{ \cancel{2}} \{2 \: \times \: 5 \: + \: (n \: - \: 1)4 \} \quad}{ \dfrac{(n \: + \: 1)}{ \cancel{2}} \{2 \: \times \: 7 \: + \: ( \: n \: + \: \cancel{1 }\: - \: \cancel{1})2 \} } \: = \: \dfrac{17}{16} \\ \\ \\ \sf \implies \dfrac{ \quad n(10 \: + \: 4n \: - \: 4) \quad}{(n \: + \: 1)(14 \: + \: 2n)} \: = \: \dfrac{17}{16} \\ \\ \\ \sf \implies \dfrac{n(4n \: + \: 6)}{(n \: + \: 1)(2n \: + \: 14)} \: = \: \dfrac{17}{16}$




$\sf \implies \dfrac{ \cancel{2}n(2n \: + \: 3)}{ \cancel{2}(n \: + \: 1)(n \: + \: 7)} \: = \: \dfrac{17}{16} \\ \\ \\ \sf \implies16n(2n \: + \: 3) \: = \: 17(n \: + 1)(n \: + \: 7) \\ \\ \\ \sf \implies32 {n}^{2} \: + \: 48n \: = \: 17( {n}^{2} \: + \: 7n \: + \: n \: + \: 7) \\ \\ \\ \sf \implies32 {n}^{2} \: + \: 48n \: = \: 17( {n}^{2} \: + \: 8n \: + \: 7) \\ \\ \\ \sf \implies32 {n}^{2} \: + \: 48n \: = \: 17 {n}^{2} \: + \: 136n \: + \: 119 \\ \\ \\ \sf \implies32 {n}^{2} \: - \: 17 {n}^{2} \: + \: 48n \: - \: 136n \: - \: 119 \: = \: 0 \\ \\ \\ \sf \implies15 {n}^{2} \: - \: 88n \: - \: 119 \: = \: 0 \\ \\ \\ \sf \implies15 {n}^{2} \: - \: 105n \: + \: 17n \: - \: 119 \: = \: 0 \\ \\ \\ \sf \implies15n(n \: - \: 7) \: + \: 17(n \: - \: 7) \: = \: 0$




$\sf \implies(n \: - \: 7)(15n \: + \: 17) \: = \: 0 \\ \\ \\ \underline{\textsf{By Zero Product Rule : }} \\ \\ \sf \implies n \: - \: 7 \: = \: 0 \qquad or \: \: \implies 15n \: + \: 17 \: = \: 0 \\ \\ \\ \sf \: \: \therefore \: \: n \: = \: 7 \: \: \qquad \qquad \: or \: \: \therefore \: \: n \: = \: - \dfrac{17}{15}$




$\textsf{Since , no. of terms can't be in negative } \\ \textsf{or fractional form . So , n = 7 .}$