Question

Find the sum of first 10 terms of an AP in which the half of the sum of first and last term is 80
​

Answer 1

$\textbf{Given:}$

$\textsf{In an A.P, half of the sum of first and last term is 80}$

$\textbf{To find:}$

$\textsf{Sum of the first 10 terms the A.P}$

$\mathsf{}$

$\textbf{Solution:}$

$\textsf{Half of the sum of first and last term is 80}$

$\implies\mathsf{\dfrac{1}{2}(a+l)=80}$

$\mathsf{Now,}$

$\textsf{Sum of the first  10 terms}$

$\mathsf{=S_{10}}$

$\mathsf{=\dfrac{n}{2}[a+l]}$

$\mathsf{=n{\times}\dfrac{1}{2}[a+l]}$

$\mathsf{=10{\times}80}$

$\mathsf{=800}$

$\mathsf{}$

$\textbf{Answer:}$

$\textsf{Sum of the  first 10 terms is 800}$

$\textbf{Find more:}$

Find the sum of all the numbers between 200 and 500 which are divisible by 7​

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Answer 2

$\textbf{Given:}$

$\textsf{In an A.P, half of the sum of first and last term is 80}$

$\textbf{To find:}$

$\textsf{Sum of the first 10 terms the A.P}$

$\mathsf{}$

$\textbf{Solution:}$

$\textsf{Half of the sum of first and last term is 80}$

$\implies\mathsf{\dfrac{1}{2}(a+l)=80}$

$\mathsf{Now,}$

$\textsf{Sum of the first  10 terms}$

$\mathsf{=S_{10}}$

$\mathsf{=\dfrac{n}{2}[a+l]}$

$\mathsf{=n{\times}\dfrac{1}{2}[a+l]}$

$\mathsf{=10{\times}80}$

$\mathsf{=800}$

$\mathsf{}$

$\textbf{Answer:}$

$\textsf{Sum of the  first 10 terms is 800}$