Question
If $\mathrm{a}^{2}+16 \mathrm{~b}^{2}+49 \mathrm{c}^{2}-4 \mathrm{ab}-7 \mathrm{ac}-28 \mathrm{bc}=0$ then $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in(a) AP(b) $\mathrm{GP}$(c) HP(d) None of these
Step 1
Step 1: Rewrite the given equation as follows: \[a^{2}-4ab+16b^{2}+7ac-28bc+49c^{2}=0\] Show more…
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