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Arithmetic Progressions — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsArithmetic Progressions2 Marks
Question
Find the sum of the following series:
$(\text{a}-\text{b})^2+(\text{a}^2+\text{b}^2)+(\text{a}+\text{b})^2+\ ...\ +[(\text{a}+\text{b}^2)+6\text{ab}]$
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Answer

$(\text{a}-\text{b})^2+(\text{a}^2+\text{b}^2)+(\text{a}+\text{b})^2+\ ...\ +[(\text{a}+\text{b}^2)+6\text{ab}]$
Let number of terms be n
Then,
$​\text{a}_\text{n}​=(\text{a}+\text{b}^2)+6\text{ab}$
$\Rightarrow(\text{a}+\text{b})^2+(\text{n}-1)(2\text{nd})=(\text{a}+\text{b})^2+6\text{ab}$
$\Rightarrow\text{a}^2+\text{b}^2-2\text{ab}+2\text{abn}-2\text{ab}=\text{a}^2+\text{b}^2+2\text{ab}+6\text{ab}$
then,
$\text{s}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$
$\text{s}_6=\frac{6}{2}[\text{a}^2+\text{b}^2-2\text{ab}+\text{a}^2+\text{b}^2+\text{ab}+6\text{ab}]$
$=6[\text{a}^2+\text{b}^2+3\text{ab}]$

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