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If $\mathrm{b}_{1}, \mathrm{~b}_{2}, \mathrm{~b}_{3}$ are three consecutive terms of an arithmetic progression with common difference $\mathrm{d}>0$, then what is the
value of $\mathrm{d}$ for which $\mathrm{b}_{3}^{2}=\mathrm{b}_{2} \mathrm{~b}_{3}+\mathrm{b}_{1} \mathrm{~d}+2 ?$
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value of $\mathrm{d}$ for which $\mathrm{b}_{3}^{2}=\mathrm{b}_{2} \mathrm{~b}_{3}+\mathrm{b}_{1} \mathrm{~d}+2 ?$
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Verified Answer
The correct answer is:
1
$b_{1}, b_{2}, b_{3}$ are in $A P$ with common difference $d$, so $b_{2}=b_{1}+d$ and $b_{3}=b_{1}+2 d$
As given, $\mathrm{b}_{3}^{2}=\mathrm{b}_{2} \mathrm{~b}_{3}+\mathrm{b}_{1} \mathrm{~d}+2$
$\Rightarrow\left(b_{1}+2 d\right)^{2}=\left(b_{1}+d\right)\left(b_{1}+2 d\right)+b_{1} d+2$
$\Rightarrow \mathrm{b}_{1}^{2}+4 \mathrm{~d}^{2}+4 \mathrm{~b}_{1} \mathrm{~d}=\mathrm{b}_{1}^{2}+2 \mathrm{~b}_{1} \mathrm{~d}+\mathrm{b}_{1} \mathrm{~d}+2 \mathrm{~d}^{2}+\mathrm{b}_{1} \mathrm{~d}+2$
$\Rightarrow 2 \mathrm{~d}^{2}=2$
$\Rightarrow \mathrm{d}^{2}=1$
$\Rightarrow \mathrm{d}=\pm 1$
i.e. $\mathrm{d}=1$ or $-1$
Since, $\mathrm{d}>0,-1$ is discarded and $\mathrm{d}=1$
As given, $\mathrm{b}_{3}^{2}=\mathrm{b}_{2} \mathrm{~b}_{3}+\mathrm{b}_{1} \mathrm{~d}+2$
$\Rightarrow\left(b_{1}+2 d\right)^{2}=\left(b_{1}+d\right)\left(b_{1}+2 d\right)+b_{1} d+2$
$\Rightarrow \mathrm{b}_{1}^{2}+4 \mathrm{~d}^{2}+4 \mathrm{~b}_{1} \mathrm{~d}=\mathrm{b}_{1}^{2}+2 \mathrm{~b}_{1} \mathrm{~d}+\mathrm{b}_{1} \mathrm{~d}+2 \mathrm{~d}^{2}+\mathrm{b}_{1} \mathrm{~d}+2$
$\Rightarrow 2 \mathrm{~d}^{2}=2$
$\Rightarrow \mathrm{d}^{2}=1$
$\Rightarrow \mathrm{d}=\pm 1$
i.e. $\mathrm{d}=1$ or $-1$
Since, $\mathrm{d}>0,-1$ is discarded and $\mathrm{d}=1$
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