Question

✌in an arithmetic progression the ratio between seventh and third term is 12:5. find the ratio between 13th and 4th term.
please answer friends ✌​

Answer 1

$\huge{\bold{ Question:-}}$

In an arithmetic progression the ratio between 7th and 3rd term is 12:5 . Find the ratio between 13th and 4th term ...

${\purple{\boxed{\large{\bold{Formula used:-}}}}}$

$nth \: term \: of \: AP\: = a + (n - 1)d \\ \\ where \: a \: is \: the \: first \: term \\ n \: is \: the \: term \: number \\ d \: is \: the \: common \: difference \: . \\ \\ \huge{\bold{ Solution:-}} \\ \\ when \: n = 7 \\ so, \: a + (7 - 1)d = a + 6d \\ \\ and \: when \: n \: = 3 \\ so, \: a + (3 - 1)d = a + 2d \\ \\ {\red{\boxed{\large{\bold{Given}}}}} \\ \\ 7th \: and \: 3rd \: term \: of \: an \: AP \: is \: 12:5 \\ \\ \frac{a + 6d}{a + 2d} = \frac{12}{5} \\ \\ 5a + 30d = 12a + 24d \\ \\ 5a - 12a = 24d - 30d \\ \\ - 7a = - 6d \\ \\ a = \frac{6d}{7} \\ \\ Here, \: 13th \: term \: = a + (13 - 1)d = a + 12d \\ \\ and \: 4th \: term \: = a + (4 - 1)d = a + 3d \\ \\ Now ,\: ratio \: of \: 13th \: and \: 4th \: term \: \\ \\ = \frac{a + 12d}{a + 3d} \\ \\ = \frac{ \frac{6d}{7} + 12d}{ \frac{6d}{7} + 3d} \\ \\ = \frac{6d + 84d}{6d + 21d} \\ \\ = \frac{90d}{27d} \\ \\ = \frac{90}{27} \\ \\ = \frac{10}{3}$

Thus , the ratio of 13th and 4th term is 10:3

${\purple{\boxed{\large{\bold{Answer = 10:3}}}}}$

Hope its helpful ❤

Answer 2

Answer:

$\large\boxed{\sf{10:3}}$

Step-by-step explanation:

Given that, in an AP,

Ratio of seventh and third term is 12:5.

To find the ratio of 13th and 4th term.

We know that, nth term of an AP is given by,

• $a_{n}=a+(n-1)d$

Where, a is first term and d is common difference.

Now, when n = 7,

$a_{7}=a+6d$

And, when n = 3,

$a_{3}=a+2d$

Therefore, according to question, we get,

$= > \dfrac{a + 6d}{a + 2d} = \dfrac{12}{5} \\ \\ = > 5(a + 6d) = 12(a + 2d) \\ \\ = > 5a + 30d = 12a + 24d \\ \\ = > 12a - 5a = 30d - 24d \\ \\ = > 7a = 6d \\ \\ = > a = \dfrac{6}{7} d$

Now, 13th and 4th term is given by (a + 12d) and (a+3d) respectively.

Therefore, the ratio will be,

$= > \dfrac{a_{ 13}}{a_{ 4 }} = \dfrac{a + 12d}{a + 3d} \\ \\ = > \dfrac{a_{ 13}}{a_{ 4 }} = \dfrac{ \dfrac{6}{7}d + 12d }{ \dfrac{6}{7}d + 3d } \\ \\ = > \dfrac{a_{ 13}}{a_{ 4 }} = \dfrac{6d + 84d}{6d + 21d} \\ \\ = > \dfrac{a_{ 13}}{a_{ 4 }} = \dfrac{90d}{27d} \\ \\ = > \dfrac{a_{ 13}}{a_{ 4 }} = \dfrac{10}{3}$

Hence, the ratio of 13th to 4th term is 10:3.