The good AP - GP

Algebra Level 5

In an increasing sequence of positive integers the first three terms are in AP. The last three terms are in GP and first and fourth terms differ by 30. Find the sum of all four terms.


The answer is 129.

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1 solution

L e t a , d , . . . . . r b e t h e u s u a l v a r i a b l e s . S o t h e i n t e g e r t e r m s a r e , a a + d a + 2 d = ( a + d ) r a + 30 = ( a + 2 d ) r . r = a + 2 d a + d = a + 30 a + 2 d ( a + 2 d ) 2 = ( a + 30 ) ( a + d ) S i m p l i f y i n g a n d s o l v i n g q u a d r a t i c i n d , d = ( 30 3 a ) ± ( 30 3 a ) 2 + 480 a 2 4 S i n c e d i s a + t i v e i n t e g e r , + ( 30 3 a ) 2 + 480 a m u s t b e a n i n t e g e r , ( 30 3 a ) 2 + 480 a = 900 + 300 a + 9 a 2 m u s t b e s q u a r e o f a n i n t e g e r . a = 18 s a t i s f i e s t h i s , ( b y t r i a l a n d e r r o r ) a n d s o l v i n g w e g e t d = 9. S o t h e t e r m s a r e 18.......27........36........48. T h e i r s u m = 129. Let~a,~d,.....r~~~be~ the ~usual ~variables. \\ So~the~~integer~terms~are,\\ a~~~~~~~a+d~~~~~~~a+2d=(a+d)*r~~~~~~~a+30=(a+2d)r.\\ \therefore~~r=\dfrac{a+2d}{a+d}=\dfrac{a+30}{a+2d}\\ \implies~(a+2d)^2=(a+30)*(a+d)\\ Simplifying ~and ~solving~quadratic~in ~d,\\ d=\dfrac{(30-3a) \pm \sqrt{(30-3a)^2+480a}}{2*4}\\ Since~d~is~a~+tive~integer,~~~+\sqrt{(30-3a)^2+480a} ~~must~be~an~integer,\\ \implies~(30-3a)^2+480a=900+300a+9a^2~~must~be~square~of~an~integer. \\ a=18~satisfies ~this,(by~trial~~and ~error)~and~solving~we~get~d=9.\\ So~the~terms~are~~~18.......27........36........48.\\ Their ~sum = \huge~~\color{#D61F06}{129}.

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