Different kind of AP.

Algebra Level 3

Let a n a_n be the n n th term of an arithmetic progression . If ∑ r = 1 1 0 99 a 2 r = 1 0 100 \displaystyle\sum_{r=1}^{10^{99}}a_{2r}=10^{100} and ∑ r = 1 1 0 99 a 2 r − 1 = 1 0 99 \displaystyle\sum_{r=1}^{10^{99}}a_{2r-1}=10^{99} , what is the common difference of the arithmetic progression?

9 1 10 7

Sahil Silare by Sahil Silare , 20, India

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2 solutions

Chew-Seong Cheong
Mar 15, 2018

Similar solution with @Vitor Juiz 's

∑ r = 1 n a 2 r = a 2 + a 4 + a 6 + ⋯ + a 2 n = 10 n where n = 1 0 99 ∑ r = 1 n a 2 r − 1 = a 1 + a 3 + a 5 + ⋯ + a 2 n − 1 = n ⟹ ∑ r = 1 n a 2 r − ∑ r = 1 n a 2 r − 1 = ( a 2 − a 1 ) + ( a 4 − a 3 ) + ( a 6 − a 5 ) + ⋯ + ( a 2 n − a 2 n − 1 ) 10 n − n = d + d + d + ⋯ + d ⏟ Number of d = n 9 n = n d ⟹ d = 9 \begin{aligned} \sum_{r=1}^n a_{2r} & = a_2 + a_4 + a_6 + \cdots + a_{2n} = 10n & \small \color{#3D99F6} \text{where }n=10^{99} \\ \sum_{r=1}^n a_{2r-1} & = a_1 + a_3 + a_5 + \cdots + a_{2n-1} = n \\ \implies \sum_{r=1}^n a_{2r} - \sum_{r=1}^n a_{2r-1} & = (a_2-a_1) + (a_4-a_3) + (a_6-a_5) + \cdots + (a_{2n}-a_{2n-1}) \\ 10n - n & = \underbrace{d + d + d + \cdots + d}_{\text{Number of }d = n} \\ 9n & = nd \\ \implies d & = \boxed{9} \end{aligned}

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Sir how do you write this kind of LaTeX? Is it practice or some software?

Sahil Silare - 3 years, 2 months ago

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I just key in the codes by hand. You can see the LaTex by placing your mouse cursor on the formulas. Or click the pull-down menu " ⋯ \cdots Menu" at the bottom of the answer section and select "Toggle LaTex". You can learn new codes using the free editor .

I learn all my LaTex after joining Brilliant.org. That is the style I have created. I am providing solutions on Brilliant practically everyday.

Chew-Seong Cheong - 3 years, 2 months ago

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Thanks sir!

Sahil Silare - 3 years, 2 months ago
Vitor Juiz
Mar 15, 2018

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