Solution: Let a and b be the roots of the quadratic equation. So, in accordance to the given condition, we have $ A.M\text{ }=\text{ }\left( a+b \right)/2\text{ }=\text{ }A $ $ a\text{ }+\text{...

Solution: It is given that A.M = 25, G.M = 20. We have, G.M = √ab and A.M = (a+b)/2 So, we can write: √ab = 20 ……. (1) (a+b)/2 = 25……. (2) Also, a + b = 50 Or, a = 50 – b Substituting the value of...

Solution: According to the question, the GM between a and b is √ab We have a = 2 and b = 1/4 Therefore, GM = √(2×1/4)= √(1/2) GM = 1/√2 Therefore, the value of a is 1/√2

(iii) –8 and –2 Solution: (iii) –8 and –2 According to the question, GM between a and b is √ab we have, a = –2 and b = –8 GM = √(–2×–8) = √–16 GM = -4

(i) 2 and 8 (ii) a3b and ab3 Solution: (i) 2 and 8 According to the question, GM between a and b is √ab We have a = 2 and b =8 GM = √2×8 = √16 GM = 4 (ii) a3b and ab3 According to the question, GM...

Solution: Suppose that the five terms are a1, a2, a3, a4, a5. where A = 32/9, B = 81/2 The above five terms are between A and B. So the GP is: A, a1, a2, a3, a4, a5, B. Therefore, there are 7 terms...

Solution: Consider the first five terms to be a1, a2, a3, a4, a5. we have, A = 27, B = 1/81 The above mentioned five terms are between A and B. So the GP is: A, a1, a2, a3, a4, a5, B. Therefore, we...

Solution: Let a1, a2, a3, a4, a5, a6 be the six terms We have, A = 27, B = 1/81 According to the question, these 6 terms are between A and B. So the GP is given by: A, a1, a2, a3, a4, a5, a6, B. So...