Consider an infinite geometric series with first term a and common ratio r. If its sum is 4 and the second term is 3/4, then (1) a = 4/7, r = 3/7... View Article
Solution: Let a and b be the two numbers. a-b = 48 (√a-√b)(√a+√b) = 48 …(i) AM = (a+b)/2 GM = √(ab) AM -GM = 18 (a+b)/2 - √(ab) = 18... View Article
Solution: Since a,b,c are in GP b2 = 4ac 4a, 5b, 4c are in AP. 2(5b) = 4a+4c 10b = 4(a+c) (a+c) = 10b/4 (a+c) = 5b/2…(i) a + b + c = 70... View Article
In a sequence of 21 terms, the first 11 terms are in AP with common difference 2 and the last 11 terms are in GP with common ratio 2. If the... View Article
Solution: Given logx ax, logx bx and logx cx are in AP. logx a +logx x, logx b +logx x, logx c +logx x are in AP. 1+logx a, 1+logx b, 1+logx... View Article
Solution: Let a and b be two numbers, then a:b = 3 + 2√2 : 3 - 2√2 AM = (a+b)/2 = ((3+ 2√2)x + (3 - 2√2)x)/2 = 6x/2 = 3x GM = √(ab) = √(3... View Article
Solution: (32) (32)1/6 (32)1/36... to ∞ = 32(1+1/6 +1/36+... Sum of infinite GP = a/1-r Here a = 1 r = 1/6 a/1-r = 1/(1-⅙) = 6/5 So... View Article
If 0<theta < pi/2, x = sigma n = 0 to infinity cos2n theta, y = sigma n = 0 to infinity sin2n theta and z = sigma n = 0 to infinity cos2n... View Article