Question

arthimatic progression

In an A.P., T3 = 8, T10 = T6 + 20. Find A.P.

Answer 1

Hope this helps......... Plz mark as brainliest....!

Answer 2

Answer:

$\large{\boxed{\red{\bf A.P.- 2, 3, 8,....}}}$

Step-by-step explanation:

Given that ;

In an Arithmetic Progression (A. P) -

t₃ = 8

t₁₀ = t₆ + 20

For t₃ = 8 :-

tₙ = a + (n - 1)d

⇒ t₃ = a + ( 3 - 1 )d

⇒ 8 = a + 2d... (i)

For t₁₀ = t₆ + 20 :-

⇒ a + 9d = a + 5d + 20

⇒ a + 9d - a - 5d = 20

⇒ 4d = 20

⇒ d = $\frac{20}{4}$

⇒ d = 5.....(ii)

Substituting the value of (ii) in (i),

8 = a + 2d

⇒ 8 = a + 2 * 5

⇒ 8 = a + 10

⇒ a = 8 - 10

⇒ a = - 2

Therefore, the terms of AP will be,

First term = a = - 2

Second term = a + d = - 2 + 5 = 3

Third term = a + 2d = - 2 + 10 = 8

Hence, the required AP is - 2, 3, 8,....tₙ