Let a1,a2,a3 be the first three terms of an A.P. such that a1+a2+a3=−12 and a1a2a3=80. Then the value of a21+a22+a23 is equal to
A
80
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B
100
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C
120
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D
160
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Solution
The correct option is C120 Let a1=a−d,a2=a,a3=a+d a1+a2+a3=−12⇒3a=−12⇒a=−4 a1a2a3=80⇒a(a2−d2)=80⇒−4(16−d2)=80⇒d=±6 ∴a1=−10,a2=−4,a3=2 or a1=2,a2=−4,a3=−10 Hence, a21+a22+a23=120