Question

the sum of n terms of an ap is (n^(2))/(4)-(25 n)/(4) .find the ap and its 15th term.​

Answer 1

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$\underline{\blacksquare\:\:\:\footnotesize{\red{\text{SolutioN:-}}}}$

$\footnotesize{\text{The given AP series' =}\:a_n=\dfrac{n^2}{4}-\dfrac{25n}{4} ;}$

$\footnotesize{\text{Where n is whole number , n=1,2,3,........}}$

$\footnotesize{\therefore\:\text{first term=}\:a_1=\dfrac{1^2}{4}-\dfrac{25\times1}{4}}$

$\footnotesize{\implies\:a_1=\dfrac{1}{4}-\dfrac{25}{4}}$

$\footnotesize{\implies\:a_1=\dfrac{1-25}{4}}$

$\footnotesize{\implies\:a_1=\dfrac{-24}{4}}$

$\footnotesize{\implies\:a_1=\dfrac{\cancel{-24}}{\cancel4}}$

$\footnotesize{\implies\:\red{a_1=-6}}$

$\footnotesize{\therefore\:\text{second term=}\:a_2=\dfrac{2^2}{4}-\dfrac{25\times2}{4}}$

$\footnotesize{\implies\:a_1=\dfrac{\cancel4}{\cancel4}-\dfrac{\cancel{50}}{\cancel4}}$

$\footnotesize{\implies\:a_2=1-\dfrac{25}{2}}$

$\footnotesize{\implies\:a_2=\dfrac{2-25}{2}}$

$\footnotesize{\implies\:a_2=\dfrac{-23}{2}}$

$\footnotesize{\implies\:\red{a_2=-11.5}}$

$\footnotesize{\therefore\:\text{common difference = d}=a_2-a_1=-11.5-(-6)=-5.5}$

$\footnotesize{\therefore\:a_3=a_2+d}$

$\footnotesize{\implies\:a_3=-11.5+(-5.5)}$

$\footnotesize{\implies\:\red{a_3=-17}}$

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$\footnotesize{\therefore\:a_{15}=a_1+(15-1)d}$

$\footnotesize{\implies\:a_{15}=-6+(14)(-5.5)}$

$\footnotesize{\implies\:a_{15}=-6-77}$

$\footnotesize{\implies\:\red{a_{15}=-83}}$

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$\footnotesize{\bf{\therefore\:\text{The series is ; -6 ,-11.5,-17,........}}}$

$\footnotesize{\bf{\therefore\:\text{The 15'th term is = -83}}}$

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