If d≠0 and a(a+d),(a+d)(a+2d),(a+2d)a are in G.P., then the common ratio is
A
2
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B
−2
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C
23
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D
−23
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Solution
The correct option is B−2 Since, a(a+d),(a+d)(a+2d),(a+2d)a are in G.P. ⇒(a+d)2(a+2d)2=a2(a+d)(a+2d) Since, (a+d)(a+2d)≠0 because no term of G.P. should be 0 ⇒(a+d)(a+2d)=a2 ⇒a2+3ad+2d2=a2 ⇒d(3a+2d)=0 ⇒3a=−2d(∵d≠0) Common ratio =a+2da=a−3aa=−2