Question

the ratio of the 3rd and 6th term of an arithmetic sequence is 4 : 5. Find the ratio of 7th and 11th terms​

Answer 1

Answer:

4 : 5

Step-by-step explanation:

Let first term be a and common difference be d.

In question,

⇒ (3rd term):(6th term) = 4:5

⇒ (a + 2d)/(a + 5d) = 4/5

⇒ 5(a + 2d) = 4(a + 5d)

⇒ 5a + 10d = 4a + 20d

⇒ 5a - 4a = 20d - 10d

⇒ a = 10d

          Therefore,

⇒ (7th term):(11th term)

⇒ (a + 6d)/(a + 10d)

⇒ (10d + 6d):(10d + 10d)

⇒ 16d:20d

⇒ 4:5

Answer 2

Given ,

the ratio of the third and sixth term of an arithmetic sequence is 4 : 5

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

$\boxed{ \tt{a_{n} = a + (n - 1)d}}$

Thus ,

(a + 2d)/(a + 5d) = 4/5

5a + 10d = 4a + 20d

a = 10d

Now , the ratio of seventh term and eleventh term will be

$\tt \implies \frac{a + 6d}{a + 10d}$

$\tt \implies \frac{10d + 6d}{10d + 10d}$

$\tt \implies \frac{16d}{20d}$

$\tt \implies\frac{4}{5}$