-1

First term $a= 0.9$, Common difference $d=0.01$ $$S_{n} = \frac n 2[2a+(n-1)d]$$ $$=\frac n 2[2(0.9)+(n-1)(0.01)]$$ $$=n(n+179)/200$$

Now my question is how we get $=n(n+179)/200$. Can explain how we get this answer. Thank you.

Shailesh
  • 3,789
changer
  • 23
  • To all the users who have given an answer, I would like to say that OP is asking how to go forward now from $n(n+179)/200$. He is not asking how can one reach there. – Vidyanshu Mishra Dec 04 '16 at 07:03
  • @THELONEWOLF. Go forward whereto? The OP asked very clearly "my question is how we get = n(n+179)/200. Can explain how we get this answer.". – dxiv Dec 04 '16 at 07:13
  • @dxiv, I agree that the language of question ask the same what you said. But does it make sense that you write the entire thing which is quite easy and then ask that how I get that – Vidyanshu Mishra Dec 04 '16 at 07:15
  • @THELONEWOLF. Sense or not, I am curious what different reading of the question you have in mind. – dxiv Dec 04 '16 at 07:17
  • @dxiv, For me it is possible to the largest extent that OP is asking how to go further from $n(n+179)/200$. As the given AP is infinite. – Vidyanshu Mishra Dec 04 '16 at 07:19
  • @THELONEWOLF. Sorry, I don't follow. Go further in what untold direction exactly? The AP is infinite, and $S_n$ will obviously tend to the same. But I see no hint in the question that it could be about asymptotic behaviors or anything related. The rule of thumb is to take questions at their face value. – dxiv Dec 04 '16 at 07:23
  • 1
    @dxiv, I respect what you think and we have our different thoughts. It is just a question which want an answer and you have given the one. Let's close it here. – Vidyanshu Mishra Dec 04 '16 at 07:26
  • @THELONEWOLF. Closed. Just to clear the air, I did not mean to dismiss your point, but I simply didn't see what else could be read, even between the lines, off the original question. Often times around here, that thing which is quite easy to you may be the very stumbling point for the OP. – dxiv Dec 04 '16 at 07:38
  • @dxiv, it was just a case of little disagreement and misunderstanding. But that's okay, it often happens. And if OP feels bad of my words I shall be sorry as I always try to avoid rudeness and be helpful for others. Thanks for mentioning that. – Vidyanshu Mishra Dec 04 '16 at 07:49
  • Okay @changer and thanks for advice. I did not posted an answer because there are already 3 answers which adress the OP. There is no need to increase the traffic – Vidyanshu Mishra Dec 04 '16 at 08:28
  • @THE LONE WOLF. Please try to solve the problem, I think your comment was not suitable. If you commented as a guide or teacher you must try to understand with hints of the question. That will only make you a perfect guide. If u comment rudely in a small topic then how will you deal big one. And i think i have written how we get which is equal to "How one can reach there". Please try to avoid such things. Thank you – changer Dec 04 '16 at 08:28

2 Answers2

0

Hint: by the sum of an arithmetic progression $a_1=0.9, a_n=a_1+ (n-1)\cdot 0.01$ so: $$S_n=n\,\frac{a_1+a_n}{2}=n \frac{0.9+0.9+(n-1)\cdot 0.01}{2} = n \frac{180 + (n-1)}{200}=\cdots$$

dxiv
  • 76,497
0

Basically we have $$ \frac{n}{2}[2(0.9) + (n-1)(0.01)]$$ $$= \frac{n}{2}[1.8 + 0.01n-0.01]$$ $$= \frac{n}{2}[1.79 + 0.01n]$$ $$=\frac{n}{2}\times \frac{1}{100}[179+n]$$ $$=\frac{n(n+179)}{200}$$ Hope it helps.