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Standard XII

Mathematics

nth Term of A.P

Let a 1, a 2,...

Question

Let $a_{1}, a_{2}, a_{3}$ be the first three terms of an A.P. such that $a_{1} + a_{2} + a_{3} = - 12$ and $a_{1} a_{2} a_{3} = 80$ . Then the value of $a_{1}^{2} + a_{2}^{2} + a_{3}^{2}$ is equal to

80

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100

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120

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160

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Solution

The correct option is C $120$
Let $a_{1} = a - d, a_{2} = a, a_{3} = a + d$
$a_{1} + a_{2} + a_{3} = - 12 \Rightarrow 3 a = - 12 \Rightarrow a = - 4$
$a_{1} a_{2} a_{3} = 80 \Rightarrow a (a^{2} - d^{2}) = 80 \Rightarrow - 4 (16 - d^{2}) = 80 \Rightarrow d = \pm 6$
$∴ a_{1} = - 10, a_{2} = - 4, a_{3} = 2$
or $a_{1} = 2, a_{2} = - 4, a_{3} = - 10$
Hence, $a_{1}^{2} + a_{2}^{2} + a_{3}^{2} = 120$

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nth Term of A.P

Standard XII Mathematics

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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

The correct option is C 120Let 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

Let $a_{1}, a_{2}, a_{3}$ be the first three terms of an A.P. such that $a_{1} + a_{2} + a_{3} = - 12$ and $a_{1} a_{2} a_{3} = 80$ . Then the value of $a_{1}^{2} + a_{2}^{2} + a_{3}^{2}$ is equal to