Question
Between 1 and $31, \mathrm{~m}$ numbers have been inserted in such a way that resulting sequence is an A.P. and the ratio of $7^{\text {th }}$ and $(m-1)^{\text {th }}$ numbers is $5: 9$. Find the value of $m$.
✓
Answer
Let $A_1, A_2, A_3, A_4$, …….., $A_m$ be $m$ numbers between 1 and 31 . Here, $a=1$ and let the common difference be d .
$\therefore a_{m+2}=31$
$\Rightarrow a+(m+2-1) d=31$
$\Rightarrow 1+(m+1) d=31$
$\Rightarrow d=\frac{30}{m+1}$
Now, $\mathrm{A}_7=a+7 d=1+7 \times\left(\frac{30}{m+1}\right)=\frac{m+1+210}{m+1}=\frac{m+211}{m+1}$
And $\mathrm{A}_{\mathrm{m}-1}=\mathrm{a}+(\mathrm{m}-1) \mathrm{d}=1+(m-1) \times\left(\frac{30}{m+1}\right)=\frac{m+1+30 m-30}{m+1}=\frac{31 m-29}{m+1}$
According to question, $\frac{A_7}{A_{m-1}}=\frac{5}{9}$
$\Rightarrow \frac{\frac{m+211}{m+1}}{\frac{31 m+29}{m+1}}=\frac{5}{9}$
$\Rightarrow \frac{m+211}{31 m-29}=\frac{5}{9}$
$\Rightarrow 9 m+1899=155 m-145$
$\Rightarrow 146 m=2044$
$\Rightarrow m=14$
$\therefore a_{m+2}=31$
$\Rightarrow a+(m+2-1) d=31$
$\Rightarrow 1+(m+1) d=31$
$\Rightarrow d=\frac{30}{m+1}$
Now, $\mathrm{A}_7=a+7 d=1+7 \times\left(\frac{30}{m+1}\right)=\frac{m+1+210}{m+1}=\frac{m+211}{m+1}$
And $\mathrm{A}_{\mathrm{m}-1}=\mathrm{a}+(\mathrm{m}-1) \mathrm{d}=1+(m-1) \times\left(\frac{30}{m+1}\right)=\frac{m+1+30 m-30}{m+1}=\frac{31 m-29}{m+1}$
According to question, $\frac{A_7}{A_{m-1}}=\frac{5}{9}$
$\Rightarrow \frac{\frac{m+211}{m+1}}{\frac{31 m+29}{m+1}}=\frac{5}{9}$
$\Rightarrow \frac{m+211}{31 m-29}=\frac{5}{9}$
$\Rightarrow 9 m+1899=155 m-145$
$\Rightarrow 146 m=2044$
$\Rightarrow m=14$
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