Question

For an increasing AP $ {{a}_{1}},{{a}_{2}},.....,{{a}_{n}}, $ if $ {{a}_{1}}+{{a}_{2}}+{{a}_{3}}=-12 $ and $ {{a}_{1}}{{a}_{3}}{{a}_{5}}=80, $ then which of the following does not hold?

• A. $ {{a}_{1}}=-10 $
• B. $ {{a}_{2}}=-1 $
• C. $ {{a}_{3}}=-4 $
• D. $ {{a}_{5}}=2 $

Answer 1

Answer: (B)

$ \because $ $ {{a}_{1}}+{{a}_{3}}+{{a}_{5}}=-12 $ $ a+(a+2d)+(a+4d)=-12 $ $ (\because d>0) $ $ \Rightarrow $ $ a+2d=-4 $ ...(i) and $ {{a}_{1}}{{a}_{3}}{{a}_{5}}=80 $ $ \Rightarrow $ $ a(a+2d)(a+4d)=80 $ $ \Rightarrow $ $ a(-4)(-4-2d+4d)=80 $ [from Eq. (i)] $ \Rightarrow $ $ (-4-2d)(-4)(-4-2d+4d)=80 $ $ \Rightarrow $ $ 4(4+2d)(-4+2d)=80 $ $ \Rightarrow $ $ 4(4{{d}^{2}}-16)=80 $ $ \Rightarrow $ $ 4{{d}^{2}}=16+20=36 $ $ \Rightarrow $ $ {{d}^{2}}=9 $ $ \therefore $ $ d=\pm 3 $ Since, AP is increasing, then $ d=+3;a=-10 $ $ \therefore $ $ {{a}_{1}}=-10;{{a}_{2}}=-7 $ $ {{a}_{3}}=a+2d=-10+6=-4 $ $ {{a}_{5}}=a+4d=-10+8=-2 $