Question

let sum of first n term of two ap are in the ratio 2n+1/3n-1 . Find ratio of there 11 th term .....plz someone tell na​

Answer 1

$\sf \to \green{{ \underline{ \underline{answer \ratio}}}} \\ \\ \sf \to \: 11th \: term \: of \: an \: ap \: = \frac{43}{62}$

$\sf \to \: \orange{{ \underline { \underline{step - \: by - \: step - \: explanation \ratio}}}}$

$\sf \to \: sum \: of \: ratio \: of \: n \: terms \: = \dfrac{2n + 1}{3n - 1} \\ \\ \sf \to \: formula \: of \: sum \: of \: n \: terms \\ \\ \sf \to \: s_{n} \: = \dfrac{n}{2} (2a \: + (n - 1)d) \\ \\ \sf \to \: \frac{ s_{n} }{ s_{n} {}^{1} } = \frac{ \dfrac{n}{2}(2a \: + (n - 1)d }{ \frac{n}{2}(2a {}^{1} + (n - 1)d {}^{1} } = \frac{2n + 1}{3n - 1} \\ \\ \sf \to \: \frac{ s_{n} }{ s_{n} {}^{1} } = \frac{2a + (n - 1)d}{2a {}^{1} + (n - 1) {d}^{1} } \\ \\ \sf \to \: \frac{ s_{n} }{ s_{n} {}^{1} } = \frac{a \: + ( \dfrac{n - 1}{2})d }{ {a}^{1} + (\dfrac{n - 1}{2} )d {}^{1} } \\ \\ \sf \to \: 11th \: term \: of \: an \: ap \: = a \: + 10d \\ \\ \sf \to \: \dfrac{ t_{11 } }{ t_{11} {}^{1} } = \dfrac{a + 10d}{ {a}^{1} + 10d {}^{1} } \\ \\ \sf \to \: \frac{n - 1}{2} = 10 \\ \\ \sf \to \: n \: = 21 \\ \\ \sf \to \: \dfrac{ t_{11} }{ t_{11 {}^{1} }} = \dfrac{a + 10d}{ {a}^{1} + 10d {}^{1} } = \: \dfrac{2n + 1}{3n - 1} \\ \\ \sf \to \: \frac{ t_{11}}{ t_{11} {}^{1} } = \frac{2 \times 21 + 1}{3 \times 21 - 1 } = \frac{43}{62} \\ \\ \sf \to \: \green{{ \underline{11th \: term \: of \: an \: ap \: in \: ratio \: = \frac{43}{62} }}}$

Answer 2

Answer:

Step-by-step explanation:

Let the two series' be T  

n

​  

 and T  

n

′

​  

 with first terms a and a  

′

 and common differences d and d  

′

 

The ratio of the sums of the series' S  

n

​  

 and S  

n

′

​  

 is given as,

S  

n

′

​  

 

S  

n

​  

 

​  

=  

[n/2][2a  

′

+[n−1]d  

′

]

[n/2][2a+[n−1]d]

​  

=  

4n+27

7n+1

​  

 

Or,

a  

′

+[(n−1)/2]d  

′

 

a+[(n−1)/2]d

​  

=  

4n+27

7n+1

​  

 ...(1)

We have to find,  

T  

11

′

​  

 

T  

11

​  

 

​  

=  

a  

′

+10d  

′

 

a+10d

​  

 

Choosing (n−1)/2=10 or n=21 in (1) we getT  

11

​  

 

​  

=  

a  

′

+10d  

′

 

a+10d

​  

=  

4(21)+27

7(21)+1

​  

=  

111

148

​  

=  

3

4

​  

.

Hence, option A.