Sequences and Series 5 Question 16
7. If $a, b, c$ are in $AP, a^{2}, b^{2}, c^{2}$ are in HP, then prove that either $a=b=c$ or $a, b,-\frac{c}{2}$ form a GP.
$(2003,4$ M)
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Answer:
Correct Answer: 7. 29
Solution:
- Since, $a, b, c$ are in an AP.
$\therefore 2 b=a+c$
and $a^{2}, b^{2}, c^{2}$ are in HP.
$$ \begin{array}{lrrr} \Rightarrow & b^{2}=\frac{2 a^{2} c^{2}}{a^{2}+c^{2}} \Rightarrow \quad \frac{a+c}{2} & =\frac{2 a^{2} c^{2}}{a^{2}+c^{2}} \\ \Rightarrow & & \left(a^{2}+c^{2}\right)\left(a^{2}+c^{2}+2 a c\right) & =8 a^{2} c^{2} \\ \Rightarrow & & \left(a^{2}+c^{2}\right)+2 a c\left(a^{2}+c^{2}\right) & =8 a^{2} c^{2} \\ \Rightarrow & & \left(a^{2}+c^{2}\right)+2 a c\left(a^{2}+c^{2}\right)+a^{2} c^{2} & =9 a^{2} c^{2} \\ \Rightarrow & & \left(a^{2}+c^{2}+a c\right)^{2}=9 a^{2} c^{2} \\ \Rightarrow & a^{2}+c^{2}+a c=3 a c \end{array} $$
$\Rightarrow \quad 4 b^{2}=-2 a c \Rightarrow b^{2}=-\frac{a c}{2}$
Hence, $a, b,-\frac{c}{2}$ are in GP.
$\therefore \quad$ Either $a=b=c$ or $a, b,-\frac{c}{2}$ are in GP.
