50 Followers Problem

Number Theory Level 2

If the sum of the terms n, 2n , 3n , of an AP are S 1 { S }_{ 1 } , S 2 { S }_{ 2 } and S 3 { S }_{ 3 } then S 3 ( S 2 − S 1 ) \frac { { S }_{ 3 } }{ { (S }_{ 2 }-{ S }_{ 1 })\quad \quad } is

0 3 1 2

Aman Baser by Aman Baser , 21, India

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2 solutions

Kalpok Guha
Dec 18, 2014

Let the first term=a,Common difference=d S 1 = 1 2 n ( 2 a + ( n − 1 ) d ) . S_1 = \frac{1}{2}n(2a+(n-1)d). S 2 = 1 2 2 n ( 2 a + ( 2 n − 1 ) d ) . S_2 = \frac{1}{2}2n(2a+(2n-1)d). S 3 = 1 2 3 n ( 2 a + ( 3 n − 1 ) d ) . S_3 = \frac{1}{2}3n(2a+(3n-1)d). Now S 2 − S 1 = 1 2 3 n ( 2 a + ( 3 n − 1 ) d ) . − 1 2 2 n ( 2 a + ( 2 n − 1 ) d ) . { S }_2 - { S }_1 =\frac{1}{2}3n(2a+(3n-1)d). - \frac{1}{2}2n(2a+(2n-1)d). By manipulating the expression we get S 2 − S 1 = n [ a + 1 2 d ( 3 n − 1 ) ] . { S }_2 - { S }_1 =n[a+\frac{1}{2}d(3n-1)]. We can write this as 1 2 n [ 2 a + d ( 3 n − 1 ) ] . \frac{1}{2}n[2a+d(3n-1)]. S 3 ( S 2 − S 1 ) = 3 \frac { { S }_{ 3 } }{ { (S }_{ 2 }-{ S }_{ 1 })\quad \quad }=3

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Chew-Seong Cheong
Dec 20, 2014

Let the first term and the common difference of the AP be a a and d d respectively. Then we have:

S 3 S 2 − S 1 = 3 n [ 2 a + ( 3 n − 1 ) d ] 2 2 n [ 2 a + ( 2 n − 1 ) d ] 2 − n [ 2 a + ( n − 1 ) d ] 2 \dfrac {S_3} {S_2-S_1} = \dfrac {\dfrac {3n[2a+(3n-1)d]} {2} } { \dfrac {2n[2a+(2n-1)d]} {2} - \dfrac {n[2a+(n-1)d]} {2}}

= 3 n [ 2 a + ( 3 n − 1 ) d ] 2 n [ 2 a + ( 2 n − 1 ) d ] − n [ 2 a + ( n − 1 ) d ] \quad \quad \quad \quad = \dfrac {3n[2a+(3n-1)d] } { 2n[2a+(2n-1)d] - n[2a+(n-1)d]}

= 3 n [ 2 a + ( 3 n − 1 ) d ] n [ 4 a + 2 ( 2 n − 1 ) d − 2 a − ( n − 1 ) d ] \quad \quad \quad \quad = \dfrac {3n[2a+(3n-1)d ]} {n[4a+2(2n-1)d - 2a-(n-1)d]}

= 3 n [ 2 a + ( 3 n − 1 ) d ] n [ 2 a + ( 4 n − 2 − n + 1 ) d ] = 3 n [ 2 a + ( 3 n − 1 ) d ] n [ 2 a + ( 3 n − 1 ) d ] = 3 \quad \quad \quad \quad = \dfrac {3n[2a+(3n-1)d ]} {n[2a+(4n-2- n+1)d]} = \dfrac {3n[2a+(3n-1)d ]} {n[2a+(3n-1)d]} = \boxed{3}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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