There are four numbers such that the first three of these form an AP and the last three a GP. If the sum of the first and third number is 2 and that of second and fourth term is 26, find the numbers.

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Solution

Dear student,

$L e t t h e f i r s t t h r e e n u m b e r s b e a_{1}, a_{1} + d a n d a_{1} + 2 d L e t t h e l a s t t h r e e n u m b e r s b e a, a r a n d a r^{2} A s t h e r e a r e t o t a l t h r e e n u m b e r s, t h u s w e c a n s a y . . . 1 s t n u m b e r = a_{1} 2 n d n u m b e r = a_{1} + d o r a 3 r d n u m b e r = a_{1} + 2 d o r a r 4 t h n u m b e r = a r^{2} T h u s w e c a n s a y, a_{1} + d = a . . . (1) a_{1} + 2 d = a r A c c . t o q u e s t i o n, s u m o f 1 s t a n d 3 r d n u m b e r = 2 \Rightarrow a_{1} + a_{1} + 2 d = 2 \Rightarrow 2 a_{1} + 2 d = 2 \Rightarrow 2 (a_{1} + d) = 2 \Rightarrow a_{1} + d = 1 . . . . (2) f r o m (1) a n d (2) w e g e t, a = 1 . . . (3) N o w, T h e l a s t 3 t e r m s a r e a, a r, a r^{2} A c c . t o q u e s t i o n s u m o f 2 n d a n d 4 t h t e r m s = 26 \Rightarrow a + a r^{2} = 26 \Rightarrow a (1 + r^{2}) = 26 f r o m (3) p u t t i n g a = 1 w e g e t, \Rightarrow (1 + r^{2}) = 26 \Rightarrow r^{2} = 26 - 1 \Rightarrow r^{2} = 25 \Rightarrow r = \sqrt{25} \Rightarrow r = \pm 5 t h u s, t h e 2 n d n u m b e r s, a = 1 3 r d n u m b e r, a r = 1 \times 5 = 5 o r 1 \times (- 5) = - 5 4 t h n u m b e r, a r^{2} = 1 \times {(\pm 5)}^{2} = 25 t h u s b y c o m p a r i n g 2 n d n u m b e r a n d 3 r d n u m b e r w e g e t, c o m m o n d i f f e r e n c e, d = a_{3} - a_{2} d = 5 - 1 = 4 o r - 5 - 1 = - 6 w e k n o w, a_{1} = a_{2} - d \Rightarrow a_{1} = 1 - 4 o r 1 - (- 6) \Rightarrow a_{1} = - 3 o r 7 t h u s t h e r e a r e t w o s e t s o f n u m b e r s w h e n d = 4 a n d a_{1} = - 3 a_{1} = - 3 a_{1} + d = a = - 3 + 4 = 1 a_{1} + 2 d = a r = - 3 + 8 = 5 a r^{2} = 1 \times {(5)}^{2} = 25 t h u s s e t f o r m i s \{- 3, 1, 5, 25\} w h e n d = - 6 a n d a_{1} = 7 a_{1} = 7 a_{1} + d = a = 7 - 6 = 1 a_{1} + 2 d = a r = 7 - 12 = - 5 a r^{2} = 1 \times {(5)}^{2} = 25 t h u s s e t f o r m i s \{7, 1, - 5, 25\}$

Regards

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