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Arithmetic Progressions — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsArithmetic Progressions4 Marks
Question
If $\text{a},\ \text{b},\ \text{c}$ are in A.P., then show that:
$\text{a}^2(\text{b}+\text{c}),\ \text{b}^2(\text{c}+\text{a}),\ \text{c}^2(\text{a}+\text{b})$ are also in A.P.
✓

Answer

$\text{a}^2(\text{b}+\text{c}),\ \text{b}^2(\text{c}+\text{a}),\ \text{c}^2(\text{a}+\text{b})$ are in A.P.
If $​​\text{b}^2(\text{c}+\text{a})-\text{a}^2(\text{b}+\text{c})=\text{c}^2(\text{a}+\text{c})$
$\Rightarrow\text{b}^2\text{c}+\text{b}^2\text{a}-\text{a}^2\text{b}-\text{a}^2\text{c}=\text{c}^2\text{a}+\text{c}^2\text{b}-\text{b}^2\text{a}-\text{b}^2\text{c}$
Given, $\text{b}-\text{a}=\text{c}-\text{b}$ $[\text{a},\ \text{b},\ \text{c}$ are inA.P.$]$
$\text{c}(\text{b}^2-\text{a}^2)+\text{ab}(\text{b}-\text{a})=\text{a}(\text{c}^2-\text{b}^2)+\text{bc}(\text{c}-\text{d})$
$(\text{b}-\text{a})(\text{ab}+\text{bc}+\text{ca})=(\text{c}-\text{b})(\text{ab}+\text{bc}+\text{ca})$
Cancelling $\text{ab}+\text{bc}+\text{ca}$ from both sides
$\text{b}-\text{a}=\text{c}-\text{b}$
$2\text{b}=\text{c}+\text{a}$ which is true
Hence, $\text{a}^2(\text{b}+\text{c}),(\text{c}+\text{a})\text{b}^2$ and $\text{c}^2(\text{a}+\text{b})$ are also in A.P.

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