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

Question

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 (A) (1/3) (B) (2/3) (C) (4/3) (D) (5/3)

Tardigrade Admin · Accepted Answer

Also, ( S a (2)/ S b (2))=(19/12)=( a 1+ a 2/ b 1+ b 2) ⇒ (7 k +4/5 k + b 2)=(19/12) ⇒ 84 k +48=95 k +19 b 2 ⇒ (48-11 k /19)= b 2....(1) Now ( S a (3)/ S b (3))=(2 a 1+2 d a /2 b 1+2 d b )=( a 1+ d a / b 1+ d b )=( a 2/ b 2) ∴ ( a 2/ b 2)=(24/14) ⇒ b 2=(7/12) × 4=(7/3) ( text As, a 2=4)...(2) Also, (7/3)=(48-11 k /19)[ operatornameusing(2). in (1) ⇒ 133=144-33 k ⇒ 33 k =11 ⇒ k =(1/3). So, a 1=(7/3) ; a 2=4 ⇒ d a =(5/3) Ans.(ii)

Q. Let $a_{n}$ and $b_{n}$ be $n^{th}$ 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

$\frac{1}{3}$

$\frac{2}{3}$

$\frac{4}{3}$

$\frac{5}{3}$

Q. Let an​ and bn​ be nth terms of two arithmetic progressions where n∈N and their sum of first n-terms be Sa​(n) and Sb​(n) respectively. Given Sb​(n)Sa​(n)​=2n+85n+9​ and a2​=4 Common difference of the sequence an​ is

31​

32​

34​

35​

Q. Let $a_{n}$ and $b_{n}$ be $n^{th}$ 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

$\frac{1}{3}$

$\frac{2}{3}$

$\frac{4}{3}$

$\frac{5}{3}$