Paragraph: Let P be the sum of first n positive terms of an...

There are two AP with the different first terms and common difference. For the first AP A: Let the first term be a and common difference $=\mathrm{d}$ Sum of n terms $\mathrm{P}=\frac{n}{2}[2 a+(n-1) d] and \mathrmn^{\text{th }}$ term $a_n=a+(n-1) d$ For the second AP B: Let the first term be M and common difference $=\mathrm{D}$ Sum of n terms $\mathrm{Q}=\frac{n}{2}[2 M+(n-1) D] and n^{\text{th }}$ term $M_n=M+(n-1) D$ It is given that $$\begin{aligned} & \frac{\text{ Sum\text{ of } } n \text{ terms of first A.P }}{\text{ Sum of } n \text{ terms of second A.P }}=\frac{5 n+4}{9 n+6} \\ & \Rightarrow \frac{\frac{n}{2}[2 a+(n-1) d]}{\frac{n}{2}[2 M+(n-1) D]}=\frac{5 n+4}{9 n+6} \\ & \Rightarrow \frac{[2 a+(n-1) d]}{[2 M+(n-1) D]}=\frac{5 n+4}{9 n+6} \\ & \Rightarrow \frac{a+\left(\frac{n-1}{2}\right) d}{M+\left(\frac{n-1}{2}\right) D}=\frac{5 n+4}{9 n+6} \ldots . . \text{ (1) } \end{aligned}$$ Now, we need to find the ratio of the first term of A and $B$$\begin{aligned} & \frac{1^{\text{st }} \text{ term\text{ of } } A}{1^{\text{st }} \text{ term of } B} \\ & =\frac{a}{M} \end{aligned}$Hence, $\frac{n-1}{2}=0 \Rightarrow n=1$ Putting $\mathrm{n}=1$ in equation (1), we get\begin{aligned} & \frac{a}{M}=\frac{5(1)+4}{9(1)+6} \\ & \Rightarrow \frac{a}{M}=\frac{9}{15} \\ & \Rightarrow \frac{a}{M}=\frac{3}{5} \\ & \Rightarrow \frac{1^{\text{st }} \text{ term of } A}{1^{\text{st }} \text{ term of } B}=\frac{3}{5} \end{aligned}Now, we need to find the ratio of the second term of $A and B\begin{aligned} & \frac{2^{\text{nd }} \text{ term of } A}{2^{\text{nd }} t \text{ erm of } B} \\ & =\frac{a+(2-1) d}{M+(2-1) D} \\ & =\frac{a+d}{M+D} \end{aligned}$ Hence, $\frac{n-1}{2}=1 \Rightarrow n=3$ Putting $\mathrm{n}=3$ in equation (1), we get\begin{aligned} & \frac{a+d}{M+D}=\frac{5(3)+4}{9(3)+6} \\ & \Rightarrow \frac{a+d}{M+D}=\frac{19}{33} \\ & \Rightarrow \frac{2^{\text{nd }} \text{ term of } A}{2^{\text{nd }} \text{ term of } B}=\frac{19}{33} \end{aligned}$$Let the first term of$A=3$and the second term of$A=19$then common difference$d=19-3=16$Let the first term of$B=5$and the second term of$B=33$then common difference$D=33-5=28$Hence,$D > d\therefore$ The correct option is (1)