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प्रश्न
If a, b, c are in continued proportion, then prove that
`a/[ a + 2b] = [a - 2b]/[ a - 4c]`
उत्तर
a, b, c are in continued proportion.
`therefore a/b = b/c = k`
⇒ `a = bk, b = ck`
⇒ `a = bk = ck xx k = ck^2`
LHS = `a/[ a + 2b] = [ck^2]/[ ck^2 + 2 xx ck ]`
= `[ck^2]/[ ck( k + 2) ]`
= `k/( k + 2)` ...(1)
RHS = `[ a - 2b]/[ a - 4c ] = [ ck^2 - 2 xx ck]/[ ck^2 - 4c ]`
= `[ck( k -2)]/[c(k^2 - 4)]`
= `[k( k -2)]/[(k+2)(k-2)]`
= `k/(k + 2)` ...(2)
From (1) and (2), we get
`a/[ a + 2b] = [a - 2b]/[ a - 4c]`
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