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Let a,b,c,d and p be any non zero distinct real numbers such that (a2 + b2 + c2)p2 – 2(ab + bc + cd)p + (b2 + c2 + d2) = 0. Then : (1) a,c,p are in G.P. (2) a,c,p are in A.P. (3) a,b,c,d are in G.P. (4) a,b,c,d are in A.P. |
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Answer» (3) a,b,c,d are in G.P. (a2 + b2 + c2)p2 + 2(ab + bc + cd)p + b2 + c2 + d2 = 0 (a2p2 + 2abp + b2) + (b2p2 + 2bcp + c2) + (c2p2 + 2cdp + d2) = 0 (ab + b)2 + (bp + c)2 + (cp + d)2 = 0 This is possible only when ap + b = 0 and bp + c = 0 and cp + d = 0 p = -b/a = -c/b = -d/c or b/a = c/b = d/c ∴ a,b,c,d are in G.P. |
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