Please answerIf a+c≠b and 1a+1c+1a-b+1c-b=0 then 1a,1b,1c are in (1)A P (2)G P (3)H P (4)none of these - Maths - Sequences and Series

Question

P l e a s e   a n s w e r I f   a + c ≠ b   a
n d   1 a + 1 c + 1 a - b + 1 c - b = 0   t h e n  
1 a , 1 b , 1 c   a r e   i n   ( 1 ) A P ( 2 ) G P
( 3 ) H P ( 4 ) n o n e   o f   t h e s e

Shruti Tyagi · Accepted Answer

Dear Student,

1a+1c+1a-b+1c-b=0⇒1a+1c-b+1c+1a-b=0⇒c-b+aac-b+a-b+cca-b=0⇒a-b+c×1ac-b+1ca-b=0since a+c≠bhence a+c-b≠0therefore 1ac-b+1ca-b=0⇒ca-cb+ac-abac×c-b×a-b=0⇒ca-cb+ac-ab=0⇒2ac-bc-ab=0⇒2ac=ab+bc⇒2b=1a+1cHence 1a,1b,1c are in AP
Regards,

P l e a s e a n s w e r I f a + c ≠ b a n d 1 a + 1 c + 1 a - b + 1 c - b = 0 t h e n 1 a , 1 b , 1 c a r e i n ( 1 ) A . P . ( 2 ) G . P . ( 3 ) H . P . ( 4 ) n o n e o f t h e s e

$P l e a s e a n s w e r I f a + c \neq b a n d \frac{1}{a} + \frac{1}{c} + \frac{1}{a - b} + \frac{1}{c - b} = 0 t h e n \frac{1}{a}, \frac{1}{b}, \frac{1}{c} a r e i n (1) A . P . (2) G . P . (3) H . P . (4) n o n e o f t h e s e$