Arithmetic Progression with 2020 terms

Algebra Level 2

The sequence $\{T_n\}$ is an arithmetic progression such that $T_2+T_{2019}=2020$ . Find the sum of its first 2020 terms, $S_{2020}$ .

The answer is 2040200.

2 solutions

Chew-Seong Cheong
Oct 14, 2020

Let the common difference of the arithmetic progression be $d$ . Then

$\begin{aligned} T_2+T_{2019} & = 2020 \\ T_1 + d + T_1 + 2018 d & = 2020 \\ T_1 + T_1 + 2019 d & = 2020 \\ T_1 + T_{2020} & = 2020 \end{aligned}$

And we have $S_n = \dfrac n2 (T_1 + T_n)$ . For $n=2020$ , $S_{2020} = \dfrac {2020}2 (T_1 + T_{2020}) = 1010 \times 2020 = \boxed{2040200}$ .

I love your way of writing solutions....Really you're solution master

SRIJAN Singh - 8 months ago

@Hypergeo H. , the word "The" in front is unnecessary, because sequence $\{T_n\}$ has not been introduced. Once introduced, we need to add "the". You should common notation such as $a_n$ instead of $T_n$ . Link reference to Brilliant's wikis, such as arithmetic progression as I have done for you, whenever possible. I am a moderator.

Chew-Seong Cheong - 8 months ago

Hi Chew-Seong, thank you for editing this problem statement and countless many others.

I see that you have removed the word "The" at the beginning of this very problem statement. That word should be there. I've added it back.

For more information, please check out this article .

Thank you so much for screening problems on Brilliant and helping to improve the community for everyone.

Brilliant Mathematics Staff - 8 months ago

Thanks for the edit @Brilliant Mathematics .

Hypergeo H. - 8 months ago

Thanks for the article on the definite article.

Chew-Seong Cheong - 8 months ago

Thanks for the edit.

Hypergeo H. - 8 months ago

You are welcome

Chew-Seong Cheong - 8 months ago

Hypergeo H.
Oct 14, 2020

$S_n=\frac n2 [{a+\ell}] = \frac n2 [{(a+1)+(\ell-1)} ]=\frac n2 [T_2+T_{n-1}] \\ S_{2020}=\frac {2020}2 [T_2+T_{2019}]=1010 (2020) = 2040200$

Note that $S_n= \frac n2 [{(a+r)+(\ell-r)} ]=\frac n2 [T_r+T_{n-1}]$ .
So the formula works as long as the indices of the two terms inside square brackets sum to $n+1$ .

Arithmetic Progression with 2020 terms

The answer is 2040200.

2 solutions

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