The average of three consecutive terms starting from the third term in

1.	The average of three consecutive terms starting from the third term in an AP is 18. What is the average of five consecutive terms of the same AP starting from the second term?1. 162. 183. 194. 20
Answer» Correct Answer - Option 2 : 18 Given: The average of three consecutive terms starting from the third term in an AP is 18. Formula: nth term of an AP, T_n= a + (n - 1)d, where ‘a’ is the first term, ‘n’ is the total number of terms and ‘d’ is the common difference. Calculation: Let the first term is ‘a’ and common difference is ‘d’. From the question (T₃ + T₄ + T₅)/3 = 18 ⇒ a + 2d + a + 3d + a + 4d = 18 × 3 ⇒ 3a + 9d = 54 ⇒ a + 3d = 18 ----(1) The average of five consecutive terms of the same AP starting from the second term is = (T₂ + T₃ + T₄ + T₅ + T₆)/5 ⇒ (a + d + a + 2d + a + 3d + a + 4d + a + 5d)/5 ⇒ (5a + 15d)/5 ⇒ a + 3d = 18 [from equation (1)] ∴ required answer is 18.

The average of three consecutive terms starting from the third term in an AP is 18. What is the average of five consecutive terms of the same AP starting from the second term?1. 162. 183. 194. 20

Answer» Correct Answer - Option 2 : 18

Given:

The average of three consecutive terms starting from the third term in an AP is 18.

Formula:

nth term of an AP, T_n= a + (n - 1)d, where ‘a’ is the first term, ‘n’ is the total number of terms and ‘d’ is the common difference.

Calculation:

Let the first term is ‘a’ and common difference is ‘d’.

From the question

(T₃ + T₄ + T₅)/3 = 18

⇒ a + 2d + a + 3d + a + 4d = 18 × 3

⇒ 3a + 9d = 54

⇒ a + 3d = 18 ----(1)

The average of five consecutive terms of the same AP starting from the second term is

= (T₂ + T₃ + T₄ + T₅ + T₆)/5

⇒ (a + d + a + 2d + a + 3d + a + 4d + a + 5d)/5

⇒ (5a + 15d)/5

⇒ a + 3d = 18 [from equation (1)]

∴ required answer is 18.