Question

The ratio of 3rd and 11th term of an A.P is 1:4. Find the ratio of its 5th term and 19th term
​

Answer 1

Given:

• Ratio of 3rd and 11th term of an AP is 1:4.

To find :

• The ratio of its 5th term and 19th term .

Solution:

$\implies \: \frac{a_3}{a_{11}} = \frac{1}{4} \\ \\ \implies \: \frac{a + 2d}{a + 10d} = \frac{1}{4} \\ \\ \implies \: 4(a + 2d) = 1(a + 10d) \\ \\ \implies \: 4a + 8d = a + 10d \\ \\ \implies \: 3a = 2d \\ \\ \implies \: a = \frac{2}{3}d$

Now ,the ratio of 5th and 19th term :

$\implies \: \frac{a_5}{a_{19}} \\ \\ \implies \: \frac{a + 4d}{a + 18d}$

Now put the value of a = 6d we get ,

$\implies \: \frac{\frac{2}{3}d + 4d}{\frac{2}{3}d + 18d} \\ \\ \implies \: \frac{\frac{14}{3}\cancel{d}}{\frac{56}{3}\cancel{d}} \\ \\ \implies \:\frac{1}{4}$

•Hence ,the ratio is 1:4.

Answer 2

GIVEN :-

• Tʜᴇ ʀᴀᴛɪᴏ ᴏғ 3ʀᴅ ᴀɴᴅ 11ᴛʜ ᴛᴇʀᴍ ᴏғ ᴀɴ A.P ɪs 1:4 .

TO FIND :-

• Tʜᴇ ʀᴀᴛɪᴏ ᴏғ ɪᴛs 5ᴛʜ ᴀɴᴅ 19ᴛʜ ᴛᴇʀᴍ .

SOLUTION :-

Wᴇ ʜᴀᴠᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

$\orange\star\:\bf{\gray{\underline{\pink{\boxed{\purple{a_n\:=\:a\:+\:(n\:-\:1)\:d\:}}}}}}$

Where,

• a = first term

• n = no. of term .

• d = common difference .

According to the question,

$\bf{\implies\:\dfrac{a_3}{a_{11}}\:=\:\dfrac{1}{4}\:}$

$\rm{\implies\:\dfrac{a\:+\:(3\:-\:1)\:d}{a\:+\:(11\:-\:1)\:d}\:=\:\dfrac{1}{4}\:}$

$\rm{\implies\:\dfrac{a\:+\:2d}{a\:+\:10d}\:=\:\dfrac{1}{4}\:}$

$\rm{\implies\:4\times{(a\:+\:2d)}\:=\:1\times{(a\:+\:10d)}\:}$

$\rm{\implies\:4a\:+\:8d\:=\:a\:+\:10d\:}$

$\rm{\implies\:4a\:-\:a\:=\:10d\:-\:8d\:}$

$\rm{\implies\:3a\:=\:2d\:}$

$\bf\green{\implies\:a\:=\:\dfrac{2}{3}d}$

Now,

$\bf{\implies\:\dfrac{a_5}{a_{19}}\:}$

$\rm{\implies\:\dfrac{a\:+\:(5\:-\:1)\:d}{a\:+\:(19\:-\:1)\:d}\:}$

$\rm{\implies\:\dfrac{a\:+\:4d}{a\:+\:18d}\:}$

$\rm{\implies\:\dfrac{\dfrac{2}{3}d\:+\:4d}{\dfrac{2}{3}d\:+\:18d}\:}$

$\rm{\implies\:\dfrac{14d/3}{56d/3}\:}$

$\rm{\implies\:\dfrac{14d}{3}\times{\dfrac{3}{56d}}\:}$

$\bf{\implies\:\dfrac{1}{4}\:}$

$\bf\pink{\implies\:\dfrac{a_5}{a_{19}}\:=\:1\::\:4\:}$