1. Tardigrade
  2. Question
  3. Mathematics
  4. Let an and bn be n text th terms of two arithmetic progressions where n ∈ N and their sum of first n-terms be Sa(n) and Sb(n) respectively. Given (Sa(n)/Sb(n))=(5 n+9/2 n+8) and a2=4 Common difference of the sequence an is

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Q. Let $a_n$ and $b_n$ be $n^{\text {th }}$ terms of two arithmetic progressions where $n \in N$ and their sum of first $n$-terms be $S_a(n)$ and $S_b(n)$ respectively. Given $\frac{S_a(n)}{S_b(n)}=\frac{5 n+9}{2 n+8}$ and $a_2=4$
Common difference of the sequence $a_n$ is

Sequences and Series

Solution:

Also, $\frac{ S _{ a }(2)}{ S _{ b }(2)}=\frac{19}{12}=\frac{ a _1+ a _2}{ b _1+ b _2}$
$\Rightarrow \frac{7 k +4}{5 k + b _2}=\frac{19}{12} \Rightarrow 84 k +48=95 k +19 b _2 \Rightarrow \frac{48-11 k }{19}= b _2$....(1)
Now $\frac{ S _{ a }(3)}{ S _{ b }(3)}=\frac{2 a _1+2 d _{ a }}{2 b _1+2 d _{ b }}=\frac{ a _1+ d _{ a }}{ b _1+ d _{ b }}=\frac{ a _2}{ b _2}$
$\therefore \frac{ a _2}{ b _2}=\frac{24}{14} \Rightarrow b _2=\frac{7}{12} \times 4=\frac{7}{3} \left(\text { As, } a _2=4\right)$...(2)
Also, $\frac{7}{3}=\frac{48-11 k }{19}\left[\operatorname{using}(2)\right.$ in (1) $\Rightarrow 133=144-33 k \Rightarrow 33 k =11 \Rightarrow k =\frac{1}{3}$.
So, $a _1=\frac{7}{3} ; a _2=4 \Rightarrow d _{ a }=\frac{5}{3}$ Ans.(ii)