Question

The ratio of the $7^{th}$ and $(m-1{)}^{th}$ Arithmetic means of the $m$ arithmetic means between $1$ and $31$ is $5:9$ then $m=$

• A. $12$
• B. $15$
• C. $14$
• D. $17$

Answer 1

The correct option is C $14$

According to question,

$\begin{array}{c}meanis,\\ \left[1,A_1,A_2,A_3,...........................A_m,31\right],\\ commondifference:\left[d=\frac{b-a}{m+1}\right]\\ \therefore d=\frac{31-1}{m-+1}=\frac{30}{m+1}\\ then,7tharithmeticmean:\\ A_7=\left(8thterm\right)=a_8=a+98-10d\\ A_7=a+7d=1+7\left(\frac{30}{m+1}\right)=\frac{m+1+210}{m+1}\\ A_7=\frac{m+211}{m+1}--------\left(i\right)\\ and,\\ A_{m-1}=\left(a_{m}term\right)=a+(m-1)d\\ =1+(m-1)\left(\frac{30}{m+1}\right)\\ =\frac{m+1+30m-30}{m+1}\\ =\frac{31m-29}{m+1}\\ A_{m-1}=\frac{31m-29}{m+1}-----------\left(ii\right)\\ Now,\\ \Rightarrow \frac{A_7}{A_{m-1}}=\frac{5}{9}\\ \Rightarrow \frac{\frac{m+211}{m+1}}{\frac{31m-29}{m+1}}=\frac{5}{9}\\ \Rightarrow \frac{m+211}{31m-29}=\frac{5}{9}\\ \Rightarrow 9m+1899=155m-145\\ \Rightarrow 146m=2044\\ \therefore m=\frac{2044}{146}=14\\ So,m=14,andthecorrectoptionisC.\end{array}$