Question

find term and common difference of an ap are 1 and 2 respectivley, find S75​

Answer 1

AnsweR:-

Sum = 5625

ExplanatioN:-

$\sf \pink{Given}\begin{cases}&\sf{First\:term\:of\:the\:AP\:is\:\bf{1.}} \\ \\ &\sf{Common\:Differnce\:of\:the\:AP\:is\:\bf{2.}}\end{cases}$

To FinD:-

The sum of the 75th term.

Solution:-

We know that,

$\normalsize\quad\quad\quad{\pink{\underline{\boxed{\bf{S_{n}=\dfrac{n}{2}\left[2a+(n-1)d\right]}}}}}$

$\sf \pink{Here}\begin{cases}&\sf{S_{n}=sum\:of\:the\:respective\:term\:of\:the\:AP=\bf{S_{75}}} \\ \\ &\sf{n=respective\:term\:of\:the\:AP=\bf{75}} \\ \\ &\sf{d=common\:differnce\:of\:the\:AP=\bf{2}} \\ \\ &\sf{a=first\:term\:of\:the\:AP=\bf{1}}\end{cases}$

Putting the values,

$\normalsize\implies{\sf{S_{75}=\dfrac{75}{2}\left[2\times1+(75-1)2\right]}}$

$\normalsize\implies{\sf{S_{75}=\dfrac{75}{2}\left[2+74\times2\right]}}$

$\normalsize\implies{\sf{S_{75}=\dfrac{75}{2}\left[2+148\right]}}$

$\normalsize\implies{\sf{S_{75}=\dfrac{75}{2}\times\left[150\right]}}$

$\normalsize\implies{\sf{S_{75}=\dfrac{75}{2}\times150}}$

$\normalsize\implies{\sf{S_{75}=\dfrac{75}{\cancel{2}}\times\cancel{150}}}$

$\normalsize\implies{\sf{S_{75}=75\times75}}$

$\large\quad\quad\quad\therefore\boxed{\mathfrak{\pink{S_{75}=5625.}}}$

The sum of the 75th term of an AP is 5625.