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Arithmetic Progressions — Maths STD 10 — Question

CBSE BoardEnglish MediumSTD 10MathsArithmetic Progressions3 Marks
Question
Find the sum:
$\frac{\text{a}-\text{b}}{\text{a}+\text{b}}+\frac{3\text{a}-2\text{b}}{\text{a}+\text{b}}+\frac{5\text{a}-3\text{b}}{\text{a}+\text{b}}+......\text{to 11 terms.}$
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Answer

Here, first term $\big(\text{A}\big)=\frac{\text{a}-\text{b}}{\text{a}+\text{b}}$
and common difference,
$\text{D}=\frac{3\text{a}-2\text{b}}{\text{a}+\text{b}}-\frac{\text{a}-\text{b}}{\text{a}+\text{b}}=\frac{\text{2a}-\text{b}}{\text{2a}+\text{b}}$
$\because$ Sum of n terms of an AP,
$\Rightarrow\text{S}_{\text{n}}=\frac{\text{n}}{2}\left\{2\frac{\big(\text{a}-\text{b}\big)}{\big(\text{a}+\text{b}\big)}+\big(\text{n}-1\big)\frac{2\text{a}-\text{b}}{\big(\text{a}+\text{b}\big)}\right\}$
$\text{S}_{\text{n}}=\frac{\text{n}}{2}\left\{\frac{2\text{a}-2\text{b}+2\text{an}-2\text{a}-\text{bn}+\text{b}}{\text{a}+\text{b}}\right\}$
$\text{S}_{\text{n}}=\frac{\text{n}}{2}\left\{\frac{2\text{an}-\text{bn}-\text{b}}{\text{a}+\text{b}}\right\}$
$\therefore$ $\text{S}_{\text{11}}=\frac{\text{11}}{2}\left\{\frac{2\text{a}\big(11\big)-\text{b}\big(11\big)-\text{b}}{\text{a}+\text{b}}\right\}$
$\text{S}_{\text{11}}=\frac{11\big(11\text{a}-6\text{b}\big)}{\text{a}+\text{b}}$
$\text{S}_{\text{11}}=\frac{11}{2}\bigg(\frac{22\text{a}-12\text{b}}{\text{a}+\text{b}}\bigg)$

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