If \( A_{1}, A_{2} \) be two \( A M \) 's and \( a \) and \( b \) , then \( \frac{A_{1}+A_{2}}{G_{1} G_{2}} \) is equal to
As per question AP series will be $a, A_1, A_2, b$ $$d = \frac{b-a}{3}$$ $$A_1 = a + \frac{b-a}{3} = \frac{2a+b}{3}$$ $$A_2 = a + \frac{2(b-a)}{3} = \frac{a+2b}{3}$$ GP series will be $a, G_1, G_2, b$ $$r = \left(\frac{b}{a}\right)^{1/3}$$ $$G_1 = a\left(\frac{b}{a}\right)^{1/3} = b^{1/3}a^{2/3}$$ $$G_2 = a\left(\frac{b}{a}\right)^{2/3} = b^{2/3}a^{1/3}$$ Required ratio will be $\frac{A_1 + A_2}{G_1 G_2} = \frac{\frac{2a+b}{3} + \frac{a+2b}{3}}{a^{2/3}b^{1/3} a^{1/3}b^{2/3}} = \frac{\frac{3a+3b}{3}}{ab} = \frac{a+b}{ab}$
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