Question

If$1,A1,A2,A3,....An,$$31$ are in $A.P.$ and $A7:An-1=5:9$ then $n$

• A. $28$
• B. $14$
• C. $15$
• D. $13$

Answer 1

The correct option is B

$14$

Explanation for correct option:

Finding the value of $n$:

Given ,$1,A_{1,}{A}_2,A_3,...A_n,31$ are in $A.P.$

So $1+\left(n+1\right)d=31$ as there are $n+2$ terms

$\Rightarrow d=\frac{30}{\left(n+1\right)}...\left(i\right)$

$$\begin{gathered} A_7:A_{n-1}=5:9 \\ \frac{\left(1+7d\right)}{\left(1+\left(n-1\right)d\right)}=\frac{5}{9} \end{gathered}$$

Substitute $d$ in above equation$\left(i\right)$

$$\begin{gathered} \frac{\left(1+7\left({\displaystyle \frac{30}{n+1}}\right)\right)}{\left(1+\left(n-1\right)\left({\displaystyle \frac{30}{n+1}}\right)\right)}=\frac{5}{9} \\ \Rightarrow \frac{\left({\displaystyle \frac{\left(n+1\right)+210}{\left(n+1\right)}}\right)}{{\displaystyle \frac{\left(\left(n+1\right)+\left(n-1\right)30\right)}{\left(n+1\right)}}}=\frac{5}{9} \\ \Rightarrow \frac{\left({\displaystyle \frac{n+211}{\left(n+1\right)}}\right)}{{\displaystyle \frac{\left(n+1\right)+30n-30}{\left(n+1\right)}}}=\frac{5}{9} \\ \Rightarrow \frac{\left(n+211\right)}{(31n-29)}=\frac{5}{9} \\ \Rightarrow \left(n+211\right)9=5(31n-29) \\ \Rightarrow 9n+1899=155n-145 \\ \Rightarrow 9n-155n+1899+145=0 \\ \Rightarrow -146n+2044=0 \\ \Rightarrow 2044=146n \\ \Rightarrow \mathit{n}\mathbf{=}\frac{\mathbf{2044}}{\mathbf{146}} \\ \Rightarrow \mathit{n}\mathbf{=}\mathbf{14} \end{gathered}$$

Hence option(B) is correct.