(i) দেওয়া আছে, a : b = b : c,
বা, \frac{a}{b} = \frac{b}{c}
\therefore b^2 = ac
বামপক্ষ = \frac{a}{c}
ডানপক্ষ = \frac{a^2 + b^2}{b^2 + c^2}
= \frac{a^2 + ac}{ac + c^2}
= \frac{a (a + c)}{c (a + c)}
= \frac{a}{c}
অর্থাৎ, \frac{a}{c} = \frac{a^2 + b^2}{b^2 + c^2}
Class 9-10 math ex 11.1 solution
(ii) দেওয়া আছে, a : b = b : c,
বা, \frac{a}{b} = \frac{b}{c}
\therefore b^2 = ac
বামপক্ষ = a^2b^2c^2 (\frac1{a^3} + \frac1{b^3} + \frac1{c^3})
= \frac{a^2b^2c^2}{a^3} + \frac{a^2b^2c^2}{b^3} + \frac{a^2b^2c^2}{c^3}
= \frac{b^2c^2}{a} + \frac{a^2c^2}{b} + \frac{a^2b^2}{c}
= \frac{ac.c^2}{a} + \frac{(b^2)^2}{b} + \frac{a^2.ac}{c} [\Since b^2 = ac]
= \frac{a.c^3}{a} + \frac{b^4}{b} + \frac{a^3c}{c}
= c^3 + b^3 + a^3
= a^3 + b^3 + c^3
= ডানপক্ষ
অর্থাৎ, a^2b^2c^2 (\frac1{a^3} + \frac1{b^3} + \frac1{c^3}) = a^3 + b^3 + c^3
(iii) দেওয়া আছে, a : b = b : c,
বা, \frac{a}{b} = \frac{b}{c}
\therefore b^2 = ac
বামপক্ষ = \frac{abc (a + b + c)^3} {(ab + bc + ca)^3}
= \frac{b. b^2 (a + b + c)^3}{(ab + bc + b^2)^3} [\because b^2 = ac]
= \frac{b^3 (a + b + c)^3}{{b (a + c + b)}^3}
= \frac{b^3 (a + b + c)^3}{b^3 (a + b + c)^3}
= 1 = ডানপক্ষ
(iv) দেওয়া আছে, a : b = b : c,
বা, \frac{a}{b} = \frac{b}{c}
\therefore b^2 = ac
১ম পক্ষ = a – 2b + c
২য় পক্ষ = \frac{(a – b)^2}a
= \frac{a^2 – 2ab + b^2}{a}
= \frac{a^2 – 2ab + ac}{a} [\since b^2 = ac]
= \frac{a(a – 2b + c)}{a}
= a – 2b + c
৩য় পক্ষ = \frac{(b – c)^2}{c}
= \frac{b^2 – 2bc + c^2}{c}
= \frac{ac – 2bc + c^2}{c} [\since b^2 = ac]
= \frac{c(a – 2b + c)}{c}
= a – 2b + c
অর্থাৎ, a – 2b + c = \frac{(a – b)^2}a = \frac{(b – c)^2}{c}
