Question

The ratio between 5th term and 8th term of an arithmetic

sequence is 5:11. Then find the ratio between 6th term

and 9th term?​

Answer 1

$Let \: 'a' \:and \: 'd' \: are \: first \:term \: and$

$Common \: difference \: of \: an \: A.P .$

/* We know that , */

$\boxed{\pink{ n^{th} \: term (a_{n}) = a + (n-1)d }}$

$Ratio \: of \: between \: 5^{th} \: term \: and$

$8^{th} \: term = 5 : 11 \: ( given )$

$\implies \frac{a_{5}}{a_{8}} = \frac{5}{11}$

$\implies \frac{a + 4d }{a + 7d } = \frac{5}{11}$

$\implies 11(a+4d) = 5(a+7d)$

$\implies 11a + 44d = 5a + 35d$

$\implies 11a - 5a = 35d - 44d$

$\implies 6a = -9d$

$\implies \frac{a}{d} = \frac{-9 }{6}$

$\implies \frac{a}{d} = \frac{-3}{2} \: --(1)$

$Now, The \: ratio \: between \: 6^{th} \:term$

$and \: 9^{th} \: term = \frac{a_{6}}{a_{9}}$

$= \frac{a + 5d }{a + 8d}$

/* Dividing numerator and denominator by d, we get */

$= \frac{\frac{a}{d} + 5 }{\frac{a}{d} + 8}$

$= \frac{ \frac{-3}{2} + 5}{\frac{-3}{2} + 8 }$

$= \frac{ \frac{-3+10}{2}}{\frac{-3+16}{2}}$

$= \frac{ 7}{13}$

$= 7 : 13$

Therefore.,

$\red{ The \: ratio \: between \: 6^{th} \:term}$

$\red{and \: 9^{th} \: term}\green { = 7: 13}$

•••♪

Answer 2

Answer:

7:13

Step-by-step explanation:

5th term : 8th term = 5:11

hence 5th term 5x and 8th term is 11x

commen difference = difference of terms/ difference of position

commen difference = (11x-5x)/8-5

= 6x/3

= 2x

6th term = 5x+2x= 7x

9th term = 11x+2x= 13x

so the ratio of 6th term and 9th term is 7x:13x = 7:13