Critical Thinking for Arithmetic Progressions

Number Theory Level 3

Given S A = 2 m + ( 2 m + 1 ) + ( 2 m + 2 ) + . . . + 4 m S_A = 2m+(2m+1)+(2m+2)+...+4m and S B = ( 2 m + 1 ) + ( 2 m + 3 ) + ( 2 m + 5 ) + . . . + ( 4 m − 1 ) S_B=(2m+1)+(2m+3)+(2m+5)+...+(4m-1) . If S A S B = k + 1 l \frac{S_A}{S_B}=k+\frac{1}{l} , find the value of k + l k+l .

2 + m 2+m 3 + m 3+m 1 + m 1+m 2 m 2m 1 + 2 m 1+2m 2 + 2 m 2+2m

Tan Chin Cheern by Tan Chin Cheern , 18, Malaysia

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1 solution

Number of terms in the first series is 2 m + 1 2m+1 . So S A = 3 m ( 2 m + 1 ) S_A=3m(2m+1) . Number of terms in the second series is m m . So S B = 3 m 2 S_B=3m^2 . Therefore S A S B = 2 m + 1 m = 2 + 1 m \dfrac{S_A}{S_B}=\dfrac{2m+1}{m}=2+\dfrac{1}{m} . So k = 2 , l = m , k + l = 2 + m k=2, l=m, k+l=\boxed {2+m}

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