logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XI  >>  Math  >>  Sequences and Series
0 votes

Between $1\:\:and\:\:31$ $m$ numbers are inserted so that the resulting sequence is an $A.P.$ and the ratio of $7^{th}\:\;and\:\:(m-1)^{th}$ numbers is $5:9$. find the value of $m$.

$\begin{array}{1 1}13 \\ 12 \\ 14 \\ 15 \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $r^{th}A.M.$ between $a\:\;and\:\:b=A_r=a+rd$ where $d=\large\frac{b-a}{n+1}$$\:\:and\:\:n$ is no. of $A.M.^s$ inserted.
Given that $m\:\:A.M.^s$ are inserted between $1\:\:and\:\:31$
$\Rightarrow\:1,A_1,A_2,A_3,....A_m,31$ are in $A.P.$
Also given that $A_7:A_{m-1}=5:9$
We know that $A_r=a+rd$, $a=1,\:b=31$
and $d=\large\frac{b-a}{m+1}$
$A_7=a+7d$ and $A_{m-1}=a+(m-1)d$
$\Rightarrow\:\large\frac{A_7}{A_{m-1}}=\frac{a+7d}{a+(m-1)d}=\frac{5}{9}$
and $d=\large\frac{31-1}{m+1}=\frac{30}{m+1}$
$\Rightarrow\:\large\frac{1+7d}{1+(m-1)d}=\frac{5}{9}$
$\Rightarrow\:9+63d=5+5(m-1)d$
$\Rightarrow\:d(68-5m)=-4$
$\large\frac{30}{m+1}$$.(68-5m)=-4$
$\Rightarrow\:30(68-5m)=-4(m+1)$
$\Rightarrow\:15(68-5m)=-2m-2$
$\Rightarrow\:73m=1022$ or $m=14$
answered Feb 24, 2014 by rvidyagovindarajan_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...