Let 3 , a , b , c be in A . P . and 3 , a − 1 , b + 1 , c + 9 be in G . P . Then, the arithmetic mean of a , b and c is:

Step-by-step solution

Given: 3 , a , b , c are in A . P So, the common difference will be d = a - 3 Now, b = 2 a - 3 and c = 3 a - 6 Also, 3 , a - 1 , b + 1 , c + 9 are in G . P So, 3 c + 9 = b + 1 a - 1 ⇒ 3 3 a - 6 + 9 = 2 a - 3 + 1 a - 1 ⇒ 3 3 a + 3 = 2 a - 2 a - 1 ⇒ 9 a + 9 = 2 a 2 - 4 a + 2 ⇒ 2 a 2 - 13 a - 7 = 0 ⇒ a = 7 , - 1 2 rejected So, b = 2 × 7 - 3 = 11 & c = 3 × 7 - 6 = 15 Now, the A . M of a , b , c is given by, ⇒ A = a + b + c 3 ⇒ A = 7 + 11 + 15 3 ⇒ A = 33 3 ⇒ A = 11