Question

in an ap series of 100 terms a1 +a50 +a51+a100=400, find the sum of all the terms of the series

Answer 1

$\textbf{Given:}$

$\textsf{In an A.P series}$

$\mathsf{a_1+a_{50}+a_{51}+a_{100}=400}$

$\textbf{To find:}$

$\textsf{Sum of all terms of the A.P series}$

$\mathsf{}$

$\textbf{Solution:}$

$\textbf{Formula used:}$

$\boxed{\begin{minipage}{7cm} \\\textsf{1. The n th term of the A.P a, a+d, a+2d.........}\\\\\mathsf{\;\;\;t_n=a+(n-1)d}\\\\\textsf{2.Sum of first n terms of the A.P is}\\\\\mathsf{\;\;S_n=\dfrac{n}{2}[2a+(n-1)d]} \end{minipage}}$

$\mathsf{Consider,}$

$\mathsf{a_1+a_{50}+a_{51}+a_{100}=400}$

$\textsf{Using formula (1),}$

$\implies\mathsf{a+(a+49d)+(a+50d)+(a+99d)=400}$

$\implies\mathsf{4a+198d=400}$

$\mathsf{Now}$

$\textsf{Sum of all the terms}$

$\mathsf{=S_{100}}$

$\textsf{Using formula (2),}$

$\mathsf{=\dfrac{100}{2}[2a+(100-1)d]}$

$\mathsf{=\dfrac{100}{2}[2a+99\,d]}$

$\mathsf{=\dfrac{50{\times}2}{2}[2a+99\,d]}$

$\mathsf{=\dfrac{50}{2}[4a+198\,d]}$

$\mathsf{=25[400]}$

$\mathsf{=10000}$

$\textbf{Answer:}$

$\textsf{Sum of all the terms of the series is 10000}$

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