Question

An arithmetic progression having its 5th and 10th in the ratio of 1:2 and T12=36. Find the arithmetic progression.

Answer 1

Answer

• 3 , 6 , 9 , 12 , ...

Given

• An arithmetic progression having its 5th and 10th in the ratio of 1:2 and T₁₂=36

To Find

• AP

Solution

A/c , " An arithmetic progression having its 5th and 10th in the ratio of 1:2 "

a₅ / a₁₀ = 1 / 2

⇒ (a + 4d) / (a + 9d) = 1 / 2

⇒ 2a + 8d = a + 9d

⇒ a = d

Also , Given T₁₂=36

⇒ a + 11d = 36

⇒ a + 11(a) = 36

⇒ 12a = 36

⇒ a = 3

So , a = d = 3 .

Now , AP is ,

a , a+d , a+2d , a+3d , a+4d , ....

⇒ a , a+a , a+2a , a+3a , a+4a , ....

⇒ a , 2a + 3a , 4a , 5a , ....

⇒ 3 , 6 , 9 , 12 , ...