Between 1 and 31, $m$ numbers have been inserted in such a way that the resulting sequence is in AP and the ratio of 7th and ${{(m-1)}^{th}}$ numbers is 5 : 9. Find the value of $m$.
ANSWER
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Hint: Here, the first number is 1 and the last number is 31, we are adding m numbers in between. Therefore total numbers are $n=m+2$. Since last term is given, find the value of d by the formula:
${{a}_{n}}=a+(n-1)d$. Now write the ratio of 7th and ${{(m-1)}^{th}}$ number which is given by: $\dfrac{a+7d}{a+(m-1)d}=\dfrac{5}{9}$. From this find the value of m.
Complete step-by-step answer:
We are given that $m$ numbers are inserted between 1 and 31 with the resulting sequence in AP. It is also given that the ratio of 7th and ${{(m-1)}^{th}}$ numbers is 5 : 9. Now, we have to find the value of $m$. Since, $m$ numbers are inserted between 1 and 31 we can say that:
Total number of terms in AP = $n=m+2$ (including 1,31 and m numbers in between) Here the first term is given as: $a=1$
The last term is given as: ${{a}_{n}}=31$
We know that the formula for the last term is: ${{a}_{n}}=a+(n-1)d$
Now, by substituting the values for $a,{{a}_{n}}$ and $n$, we get: $\begin{align} & 31=1+(m+2-1)d \\ & 31=1+(m+1)d \\ \end{align}$
Next, by taking 1 to the left side, 1 becomes -1, we obtain: $\begin{align} & 31-1=(m+1)d \\ & 30=(m+1)d \\ \end{align}$
Now, by cross multiplication we get the value of d, which is given by: $d=\dfrac{30}{m+1}$ ….. (1) The ratios of 7th term and ${{(m-1)}^{th}}$ number is given as 5 : 9
We know that the 7th number is the 8th term and is given as: $\begin{align} & {{a}_{8}}=a+(8-1)d \\ & {{a}_{8}}=1+7d \\ \end{align}$.
Similarly, ${{(m-1)}^{th}}$ number is the ${{m}^{th}}$ term and is given as: $\begin{align} & {{a}_{m}}=a+(m-1)d \\ & {{a}_{m}}=1+(m-1)d \\ \end{align}$ Hence, we can say that: $\dfrac{1+7d}{1+(m-1)d}=\dfrac{5}{9}$ By cross multiplication we get: $\begin{align} & 9(1+7d)=5(1+(m-1)d) \\ & 9+63d=5+5(m-1)d \\ \end{align}$
Now, take constants to one side and variables to the other side, 5 to the left side becomes -5 and 63d goes to the right side and becomes -63. Hence, we get: $\begin{align} & 9-5=5(m-1)d-63d \\ & 4=5d(m-1)-63d \\ \end{align}$
On the right side d is the common factor, so take outside d, we get: $\begin{align} & 4=d(5(m-1)-63) \\ & 4=d(5m-5-63) \\ & 4=d(5m-68) \\ \end{align}$
Now, substitute the value of d in equation (1) we obtain: $4=\dfrac{30}{m+1}(5m-68)$ Next, by cross multiplication we get: $\begin{align} & 4(m+1)=30(5m-68) \\ & 4\times m+4\times 1=30\times 5m+30\times -68 \\ & 4m+4=150m-2040 \\ \end{align}$
Now, take constants to one side and variables to the other, -2040 goes to left side and becomes 2040, $4m$ goes to right side and becomes $-4m$. $\begin{align} & 4+2040=150m-4m \\ & 2044=146m \\ \end{align}$ Next, by cross multiplication we obtain: $\dfrac{2044}{146}=m$ Hence, by cancellation we get: $m=14$ Therefore, the value of $m=14$.
Note: Here, we are adding m numbers in between 1 and 31. So, while taking the ratio the 7th number and ${{(m-1)}^{th}}$numbers are taken among m numbers. Hence in AP they are 8th term and ${{m}^{th}}$ term. So don’t get confused that 7th and ${{(m-1)}^{th}}$ numbers are the 7th term and ${{(m-1)}^{th}}$ term, it may lead to wrong answers.