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Arithmetic Progressions — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSArithmetic Progressions4 Marks
Question
If $\frac{\text{b}+\text{c}}{\text{a}},\ \frac{\text{c}+\text{a}}{\text{b}},\ \frac{\text{a}+\text{b}}{b}$ are A.P., prove that: $\frac{1}{\text{a}},\ \frac{1}{\text{b}},\ \frac{1}{\text{c}}$ are in A.P.
✓

Answer

If $\frac{1}{\text{a}},\frac{1}{\text{b}},\frac{1}{\text{c}}$ are in A.P $\frac{1}{\text{b}}-\frac{1}{\text{a}}=\frac{1}{\text{c}}-\frac{1}{\text{b}}$ $\text{L.H.S}=\frac{1}{\text{b}}-\frac{1}{\text{a}}$ $=\frac{\text{a}-\text{b}}{\text{ab}}$ $=\frac{\text{a}(\text{a}-\text{b})}{\text{abc}}\ ...(\text{i})$ $\text{R.H.S}=\frac{1}{\text{c}}-\frac{1}{\text{b}}$ $=\frac{\text{a}(\text{b}-\text{c})}{\text{abc}}\ ...(\text{ii})$ Since, $\frac{\text{b}+\text{c}}{\text{a}},\frac{\text{c}+\text{a}}{\text{b}},\frac{\text{a}+\text{b}}{\text{c}}$ are in A.P $\frac{\text{b}+\text{c}}{\text{a}}-\frac{\text{c}+\text{a}}{\text{b}}=\frac{\text{c}+\text{a}}{\text{b}}-\frac{\text{a}+\text{b}}{\text{c}}$ $\frac{\text{b}^2+\text{cd}-\text{ac}-\text{a}^2}{\text{ab}}=\frac{\text{c}^2+\text{ac}-\text{ab}-\text{b}^2}{\text{bc}}$ $\Rightarrow\frac{(\text{b}-\text{a})(\text{a}+\text{b}+\text{c})}{\text{ab}}=\frac{(\text{c}-\text{d})(\text{a}+\text{b}+\text{c})}{\text{bc}}$ or $\frac{\text{a}(\text{a}-\text{c})}{\text{abc}}=\frac{\text{c(a}-\text{b})}{\text{abc}}\ .....(3)$ From (1), (2) and (3) $\text{LHS=RHS}$ Hence, $\frac{1}{\text{a}},\ \frac{1}{\text{b}},\ \frac{1}{\text{c}}$ are in A.P

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