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Question

If a, b, c, d are in G.P., prove that:

(i) (a2 + b2), (b2 + c2), (c2 + d2) are in G.P.

(ii) (a2 − b2), (b2 − c2), (c2 − d2) are in G.P.

(iii) 1a2+b2,1b2-c2,1c2+d2 are in G.P.

(iv) (a2 + b2 + c2), (ab + bc + cd), (b2 + c2 + d2) are in G.P.

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Solution

a, b, c and d are in G.P.

b2=acad=bc c2=bd .......(1)

(i) b2+c22=b22+2b2c2+c22b2+c22=ac2+b2c2+b2c2+bd2 Using (1)b2+c22=a2c2+a2d2+b2c2+b2d2 Using (1)b2+c22=a2c2+d2+b2c2+d2 b2+c22=a2+b2c2+d2Therefore, a2+b2, c2+d2 and b2+c2 are also in G.P.

(ii) b2-c22=b22-2b2c2+c22b2-c22=ac2-b2c2-b2c2+bd2 Using (1)b2-c22=a2c2-b2c2-a2d2+b2d2 Using (1)b2-c22=c2a2-b2-d2a2-b2b2-c22=a2-b2c2-d2Therefore, a2-b2, b2-c2 and c2-d2 are also in G.P.

(iii) 1b2+c22=1b22+2b2c2+1c221b2+c22=1ac2+1b2c2+1b2c2+1bd2 Using (1)1b2+c22=1a2c2+1a2d2+1b2c2+1b2d2 Using (1)1b2+c22=1a21c2+1d2+1b21c2+1d21b2+c22=1a2+b21c2+1d2Therefore, 1b2+c2, 1c2+d2 and 1b2+c2 are also in G.P.

(iv) ab+bc+cd2=ab2+bc2+cd2+2ab2c+2bc2d+2abcdab+bc+cd2=a2b2+b2c2+c2d2+ab2c+ab2c+bc2d+bc2d+abcd+abcdab+bc+cd2=a2b2+b2c2+c2d2+b2b2+acac+c2c2+bdbd+bcbc+adad Using (1)ab+bc+cd2=a2b2+a2c2+a2d2+b4+b2c2+b2d2+c2b2+c4+c2d2ab+bc+cd2=a2b2+c2+d2+b2b2+c2+d2+c2b2+c2+d2ab+bc+cd2=b2+c2+d2a2+b2+c2Therefore, a2+b2+c2, ab+bc+cd and b2+c2+d2 are also in G.P.

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