We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+3
337
2
avatar+115 

Find the sum of the terms from the (n + 1) th to the m th term inclusive of an arithmetical progression whose

first term is a and whose second term is b.

If m = 13, n =   3 and the sum is 12a, find the ratio b: a.

 

Answer (m-n){a+1/2(m+n-1)(b-a)} ; 77/75

 Feb 8, 2018
edited by OldTimer  Feb 8, 2018
 #1
avatar+21991 
+1

Find the sum of the terms from the (n + 1) th to the m th term inclusive of an arithmetical progression whose

first term is a and whose second term is b.

If m = 13, n =   3 and the sum is 12a, find the ratio b: a.

 

Answer (m-n){a+1/2(m+n-1)(b-a)} ; 77/75

 

\(\text{Formula arithmetical progression: }\\ \begin{array}{|rcll|} \hline a_1 &=& a \\ a_2 &=& a+d \\ \hline \end{array} \)

 

\(\text{second term is b: }\\ \begin{array}{|rcll|} \hline a_2 &=& a+d \\ a_2 &=& b \\ \hline a+d &=& b \\ \mathbf{d} & \mathbf{=} & \mathbf{b-a} \\ \hline \end{array}\)

 

The common difference is \(\mathbf{b-a}\)

 

\(\text{The sum of the member is: }\\ \begin{array}{|rcll|} \hline s_n = n\cdot a + \dfrac{n(n-1)}{2}\cdot d \\ \hline \end{array} \)

 

\(\text{The sum of the terms from the (n + 1) th to the m th term is: }\\ \begin{array}{|rcll|} \hline s_m-s_n &=& \underbrace{m\cdot a + \dfrac{m(m-1)}{2}\cdot (b-a)}_{=s_m} - \left[ \underbrace{n\cdot a + \dfrac{n(n-1)}{2}\cdot (b-a)}_{=s_n} \right] \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)[ m(m-1)-n(n-1) ] \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)( m^2-m-n^2+n ) \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)( m^2-n^2-m+n ) \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)[ m^2-n^2-(m-n) ] \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)[ (m-n)(m+n)-(m-n) ] \\\\ &=& a(m-n) + \left(\dfrac{b-a}{2}\right)(m-n)( m+n-1 ) \\\\ \mathbf{s_m-s_n} &\mathbf{=}& \mathbf{ (m-n)\left[ a + \left(\dfrac{b-a}{2}\right)( m+n-1 ) \right] } \\ \hline \end{array}\)

 

\(\small{ \text{If $m = 13, n = 3$ and the sum is $12a$, find the ratio $b: a$ }\\ \begin{array}{|rcll|} \hline \mathbf{s_m-s_n} &\mathbf{=}& \mathbf{ (m-n)\left[ a + \left(\dfrac{b-a}{2}\right)( m+n-1 ) \right] } \\\\ 12a & = & (m-n)\left[ a + \left(\dfrac{b-a}{2}\right)( m+n-1 ) \right] \\\\ \dfrac{12a}{m-n} & = & a + \left(\dfrac{b-a}{2}\right)( m+n-1 ) \\\\ \dfrac{12a}{m-n} & = & a + \left(\dfrac{b}{2}\right)( m+n-1 )- \left(\dfrac{a}{2}\right)( m+n-1 ) \\\\ a\left( \dfrac{12}{m-n} - 1+ \dfrac{m+n-1}{2} \right) & = & b\left( \dfrac{m+n-1}{2} \right) \qquad m = 13 \qquad n = 3 \\\\ a\left( \dfrac{12}{10} - 1+ \dfrac{15}{2} \right) & = & b\left( \dfrac{15}{2} \right) \\\\ \dfrac{b}{a} &=& \left( \dfrac{2}{15} \right)\left( \dfrac{12}{10} - 1+ \dfrac{15}{2} \right) \\\\ &=& \left( \dfrac{2}{15} \right)\left( \dfrac{12-10+75}{10} \right) \\\\ &=& \dfrac{2\cdot 77}{150} \\\\ &=& \dfrac{2\cdot 77}{2\cdot 75} \\\\ \mathbf{\dfrac{b}{a}} &\mathbf{ =}& \mathbf{ \dfrac{77}{75}} \\ \hline \end{array} } \)

 

 

laugh

 Feb 8, 2018
 #2
avatar
+2

12a = (m-n){a+1/2(m+n-1)(b-a)} , where m=13, n = 3

 

12a =(13 - 3)* [a + 1/2(13+3 -1)*(b -a)]

12a = 10*[a + 1/2(15)*(b - a)]

12a = 10*[a + 1/2(15b - 15a)]

12a = 10a + 5(15b - 15a)

12a = 10a + 75b - 75a

12a - 10a + 75a = 75b

77a = 75b

b =77a / 75, or

b/a = 77 / 75

 Feb 8, 2018

17 Online Users