Pratically possible

Algebra Level 2

Suppose that { a n } \{a_n\} is an arithmetic progression with a 1 + a 2 + a 3 + ⋯ + a 100 = 100 a_1+a_2+a_3+\cdots + a_{100} = 100 and a 101 + a 102 + a 103 + ⋯ + a 200 = 200 a_{101}+a_{102}+a_{103}+ \cdots + a_{200} = 200 . What is the value of a 2 − a 1 a_2-a_1 ?


The answer is 0.01.

Sumukh Bansal by Sumukh Bansal , 16, India

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1 solution

Maximos Stratis
Nov 8, 2017

If d d is the difference between consecutive terms then we have:
a n = a 1 + ( n − 1 ) d a_n=a_1+(n-1)d
If S n S_n is the sum of the first n n terms then we know that:
S n = n 2 ( a 1 + a n ) = n 2 ( 2 a 1 + ( n − 1 ) d S_n=\frac{n}{2}(a_1+a_n)=\frac{n}{2}(2a_1+(n-1)d
From the hypothesis we have: S 100 = 100 S_{100}=100 , S 200 − S 100 = 200 S_{200}-S_{100}=200 or S 200 = 300 S_{200}=300
In the sum equation for n = 100 n=100 we get:
S 100 = 50 ( 2 a 1 + 99 d ) ⇒ 2 a 1 + 99 d = 2 S_{100}=50(2a_1+99d)\Rightarrow 2a_1+99d=2
For n = 200 n=200 we get:
S 200 = 100 ( 2 a 1 + 199 d ) ⇒ 2 a 1 + 199 d = 3 S_{200}=100(2a_1+199d)\Rightarrow 2a_1+199d=3
By compairing the two equation we find that:
100 d = 1 ⇒ d = 1 100 100d=1\Rightarrow d=\frac{1}{100}
Therefore: a 2 − a 1 = d = 1 100 = 0.01 a_2-a_1=d=\frac{1}{100}=0.01


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