Question

The sum of first n terms of two APs are in the ratio ( 3n + 8) : ( 7n + 15). Find the ratio of their 12th terms.

Answer 1

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


$\underline{\mathsf{For \: 1st \: A.P. }} \\ \\ \\ \textsf{Given,} \\ \\ \mathsf{\implies No. \: of \:terms \: = \: n } \\ \\ \textsf{Let,} \\ \\ \mathsf{\implies First \: term \: = \: a_{1}} \\ \\ \mathsf{\implies Common \: difference \: = \: d_{1}}$



$\textsf{Using Formula : } \\ \\ \boxed{\mathsf{\implies S_{n} \: = \: \dfrac{n}{2}[2a \: + \: ( n \: - \: 1)d]}}$



$\mathsf{\implies( S_{n})_{1}\: = \: \dfrac{n}{2}[2a_{1} \: + \: ( n \: - \: 1)d_{1}]}$




$\underline{\mathsf{For \: 2nd \: A.P. }} \\ \\ \\ \textsf{Given,} \\ \\ \mathsf{\implies No. \: of \: terms\: = \: n } \\ \\ \textsf{Let,} \\ \\ \mathsf{\implies First \: term \: = \: a_{2}} \\ \\ \mathsf{\implies Common \: difference \: = \: d_{2}}$



$\textsf{Using Formula : } \\ \\ \boxed{\mathsf{\implies S_{n} \: = \: \dfrac{n}{2}[2a \: + \: ( n \: - \: 1)d]}}$



$\mathsf{\implies( S_{n})_{2} \: = \: \dfrac{n}{2}[2a_{2} \: + \: ( n \: - \: 1)d_{2}]}$



$\underline{\textsf{According to question : }}$



$\mathsf{\implies \dfrac{ (S_{n})_{1}}{(S_{n})_{2}} \: = \: \dfrac{3n \: + \: 8 }{7n \: + \: 15}}$



$\mathsf{\implies \dfrac{ \quad \cancel{\dfrac{n}{2}}[2a_{1} \: + \: (n \: - \: 1)d_{1}] \quad}{ \cancel{\dfrac{n}{2}}[2a_{2} \: + \: ( n \: - \: 1)d_{2}]} \: = \: \dfrac{3n \: + \: 8}{7n \: + \: 15}}$



$\textsf{Substitute n = 23 : } \\ \\ \\ \mathsf{\implies \dfrac{2a_{1} \: + \: ( 23 \: - \: 1)d_{1}}{ 2a_{2} \: + \: ( 23 \: - \: 1 )d_{2}} \: = \: \dfrac{3 \: \times \: 23 \: + \: 8}{ 7 \: \times \: 23 \: + \: 15 }}$



$\mathsf{\implies \dfrac{ 2a_{1} \: + \: 22d_{1}}{2a_{2} \: + \: 22 d_{2} } \: = \: \dfrac{69 \: + \: 8}{161 \: + \: 15}}$




$\mathsf{\implies \dfrac{ \cancel{2}(a_{1} \: + \: 11d_{1})}{ \cancel{2}(a_{2} \: + \: 11 d_{2})} \: = \: \dfrac{77}{176}} \\ \\ \\ \mathsf{ \implies \dfrac{a_{1} \: + \: (12 \: - \: 1)d_{1}}{a_{2} \: + \: (12 \: - \: 1)d_{2}} \: = \: \dfrac{7}{16}}$




$\mathsf{\implies \dfrac{(T_{12} )_{1}}{(T_{12})_{2}} \: = \: \dfrac{7}{16}}$



$\mathsf{\therefore \quad (T_{12})_{1} \: : \: (T_{12})_{2} \: = \: 7 \: : \: 16}$

Answer 2

Let to be first term'a'and Common difference 'd' to its Arithmetic Sequence or Progression;

According to the Question;-

$\fbox{sn = \frac{n}{2}(2a + (n - 1)d}$

Let a1, a2 be the first terms and d1, d2 be the common difference of the two Arithmetic Sequence or Progression;-

Let Sn and Sn' are the Sum of nth terms;

Sum of nth terms which are given by Formula of Summation;

$Sn = \frac{n}{2} (2a \tiny{1} + \bold{ (n - 1)d1} \\ \\ and \\ \\ \\ Sn = \frac{n}{2} (2a \tiny2+ \bold{(n - 1)d \tiny{2}} \\ \\ \\ \\ \\ \\ \implies \: \frac{Sn}{ Sn} = \frac{\frac{n}{2} (2a \tiny{1} + \bold{ (n - 1)d1}}{\frac{n}{2} (2a \tiny2+ \bold{(n - 1)d \tiny{2}}} \\ \\ \\ \implies \: \frac{Sn}{ Sn} = \frac{\frac{n}{2} (2a \tiny{1} + \bold{ (n - 1)d1}}{\frac{n}{2} (2a \tiny2+ \bold{(n - 1)d \tiny{2}}} \\ \\ \\ \\ \\ \\ \\ \\ \implies \frac{Sn}{Sn} = \frac{2 \tiny{a1 } + (n - 1)d{ \tiny1}}{2 \tiny{a2 } + (n - 1)d{ \tiny2}} \\ \\ \\ \implies \frac{Sn}{Sn} = \frac{2 \tiny{a1 } + (n - 1)d{ \tiny1}}{2 \tiny{a2 } + (n - 1)d{ \tiny2}} = \frac{3n + 8}{7n + 15} \\ \\ \\ \\ \\ \implies\frac{2 \tiny{a1 } + (n - 1)d{ \tiny1}}{2 \tiny{a2 } + (n - 1)d{ \tiny2}} = \frac{3n + 8}{7n + 15} \\ \\ \\ \\ \\ \implies \frac{a1 + ( \frac{n - 1}{2})d1 }{a2 + ( \frac{n - 1}{2})d2} = \frac{3n + 8}{7n + 15} \\ \\ \\ \\ \implies we \: know \: \: that \: \: {12}^{th} terms = a + 11d \\ \\ \\ \\ \implies taking \frac{n - 1}{2} = 11 \\ \\ \\ \\ \rightarrow{so} \: nth = 23 \\ \\ \\ \\ \implies\frac{ \tiny{a1} + 11d{ \tiny1}}{ \tiny{a2 } + 11d{ \tiny2}} = \frac{3(23) + 8}{7(23) + 15} \\ \\ \\ \implies\frac{ \tiny{a1} + 11d{ \tiny1}}{ \tiny{a2 } + 11d{ \tiny2}} = \frac{77}{176}$

$\implies\frac{ \tiny{a1} + 11d{ \tiny1}}{ \tiny{a2 } + 11d{ \tiny2}} = \frac{7}{16}$

Hence, The ratio of 12th terms of the two Arithmetic Sequence or Progression= 7:16