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Question
If a, b, c, d are in proportion, then prove that
`[11a^2 + 9ac]/[ 11b^2 + 9bd] = [ a^2 + 3ac]/[ b^2 + 3bd]`
Solution
It is given that a, b, c, d are in proportion.
`therefore a/b = c/d = k`
⇒ a = bk, c = dk
`∴ [ 11a^2 + 9ac]/[ 11b^2 + 9bd]`
= `[11(bk)^2 + 9 xx bk xx dk]/[ 11b^2 +9bd]`
= `[k^2(11b^2 + 9bd)]/[11b^2 + 9bd]`
= `k^2` ...(1)
`∴ [a^2 + 3ac]/[ b^2 + 3bd]`
= `[(bk)^2 + 3 xx bk xx dk]/[ b^2 + 3bd]`
= `[k^2( b^2 + 3bd)]/[b^2 + 3bd ]`
= `k^2` ...(2)
From (1) and (2), we get
`[ 11a^2 + 9ac]/[ 11b^2 + 9bd] = [ a^2 + 3ac]/[ b^2 + 3bd]`
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