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 Solution For arithemetic progression of four terms, addition of end terms is 14 and multiplication of middle two terms is 45 , then find that numbersSolution:Your problem -> For arithemetic progression of four terms, addition of end terms is 14 and multiplication of middle two terms is 45 , then find that numbersLet the terms are a-3d, a-d, a+d, a+3dAddition of last 2 term is 14=> (a-3d) + (a+3d) = 14=> 2a = 14=> a = 14/2 = 7Now multiplication of middle two terms is 45=> (a-d) (a+d) = 45=> a^2 - d^2 = 45=> d^2 = a^2 - 45=> d^2 = 7^2 - 45=> d^2 = 49 - 45=> d^2 = 4=> d = +- 2d = +2 => Required terms : 7-3(2), 7-2, 7+2, 7+3(2) => 1, 5, 9, 13d = -2 => Required terms : 7-3(-2), 7-(-2), 7-2, 7-3(2) => 13, 9, 5, 1 Solution provided by AtoZmath.com Any wrong solution, solution improvement, feedback then Submit Here Want to know about AtoZmath.com and me

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