(iii) a2 + b2, ab + bc, b2 + c2 Solution: (iii) a2 + b2, ab + bc, b2 + c2 According to the question, a, b, c are in GP. Making use of the property of geometric mean, b2 = ac a2 + b2, ab + bc, b2 +...

(i) a2, b2, c2 (ii) a3, b3, c3 Solution: (i) a2, b2, c2 According to the question, a, b, c are in GP. Making use of the property of geometric mean, we can write: b2 = ac Upon squaring both the sides...

(iii) (b + c) (b + d) = (c + a) (c + d) Solution: (iii) (b + c) (b + d) = (c + a) (c + d) According to the question, a, b, c are in GP. Making use of the property of geometric mean, we can write:...

(i) (ab – cd) / (b2 – c2) = (a + c) / b (ii) (a + b + c + d)2 = (a + b)2 + 2(b + c)2 + (c + d)2 Solution: (i) (ab – cd) / (b2 – c2) = (a + c) / b According to the question a, b, c are in GP. Making...

(v) (a + 2b + 2c) (a – 2b + 2c) = a2 + 4c2 Solution: (v) (a + 2b + 2c) (a – 2b + 2c) = a2 + 4c2 According to the question, a, b, c are in GP. Making use of the property of geometric mean, b2 = ac...

(iii) (a+b+c)2 / (a2 + b2 + c2) = (a+b+c) / (a-b+c) (iv) 1/(a2 – b2) + 1/b2 = 1/(b2 – c2) Solution: (iii) (a+b+c)2 / (a2 + b2 + c2) = (a+b+c) / (a-b+c) According to the question, a, b, c are in GP....

(i) a(b2 + c2) = c(a2 + b2) (ii) a2b2c2 [1/a3 + 1/b3 + 1/c3] = a3 + b3 + c3 Solution: (i) a(b2 + c2) = c(a2 + b2) According to the question, a, b, c are in GP. Making use of the property of...

Solution: Suppose that the three numbers are a, ar, ar2 According to the question statement, a + ar + ar2 = 56 … (1) subtracting 1,7,21 respectively from these terms,we get: (a – 1), (ar – 7),...

Solution: Consider the first term of an A.P. to be ‘a’ and let its common difference be given by ‘d’. Therefore, we can write: b = a + d and c = a + 2d. It is gien that: a + b + c = 18 Or, we can...

 Solution: Consider the first term of an A.P. to be ‘a’ and let its common difference be‘d’. Therefore, we can write: a1 + a2 + a3 = 21 Where the three numbers are as follows: a, a + d, and a + 2d...

Solution: Consider the first term of an A.P. to be ‘a’ and let its common difference be‘d’. We have, a1 + a2 + a3 = 15 Where the three numbers are as follows: a, a + d, and a + 2d So, we can write:...

Solution: Suppose that a = k + 9; b = k−6; and c = 4; According to the question, a, b and c are in GP, then Making use of the property of geometric mean, we get: b2 = ac $ {{\left( k-6...

Solution: According to the question, a, b and c are in GP Making use of the property of geometric mean, we can write: b2 = ac Taking log on both sides with base m, we get: $...

Solution: According to the question, a, b and c are in G.P. By making use of the property of Geometric mean, we can write: $ {{b}^{2}}=~ac $ $ ~{{\left( {{b}^{2}} \right)}^{n}}~=~{{\left( ac...