Please answer as soon as possible 99 (i) I a, b, c are in - Maths - Arithmetic Progressions

Question

Please answer as soon as possible

99   ( i )       I   a ,   b ,  
c   a r e   i n   A P ,   t h e n   s h o
w   t h a t   1 b   +   c ,   1 c   +
  a ,   1 a   +   b   a r e   i n
  A P ( i i )   b   +   c   -   c a ,
  c   +   a   -   b b   a   +
  b   -   c c   a r e   i n   A P
  p r o v e   t h a t   1 a ,   1 b ,   1
c   a r e   i n   A P

Shruti Tyagi · Accepted Answer

Dear Student,

i) a,b,c are in APHence b-a=c-b⇒2b=a+c  --1to prove 1b+c,1a+c,1b+a are in APwe should have 1a+c-1b+c=1b+a-1a+c⇒b+c-a-ca+cb+c=a+c-b-ab+aa+c⇒b-aa+cb+c=c-bb+aa+c⇒b-ab+c=c-bb+a⇒b+a×b-a=b+c×c-b⇒b2-a2=c2-b2⇒b-a=c-b⇒2b=a+cas we know from equation 1 that a,b,c are in AP therefore 2b=a+cHence 1b+c,1a+c,1b+a are in AP
ii) b+c-aa,c+a-bb,a+b-cc are in APhence c+a-bb- b+c-aa=a+b-cc-c+a-bb⇒ac+a2-ab-b2-bc+abab=ab+b2-bc-c2-ac+bcbc⇒ac+a2-b2-bcab=ab+b2-c2-acbc⇒ac-bc+a-ba+bab=ab-ac+b-cb+cbc  --using a2-b2=a-ba+b ⇒ca-b+a-ba+bab=ab-c+b-cb+cbc⇒a-ba+b+ca=b-cb+c+ac⇒a-ba=b-cc⇒ac-bc=ab-acdivide by abc⇒ac-bcabc=ab-acabc⇒1b-1a=1c-1bHence 1a,1b,1c are in AP

Regards,

Please answer as soon as possible. 99 . ( i ) I a , b , c a r e i n A . P . , t h e n s h o w t h a t 1 b + c , 1 c + a , 1 a + b a r e i n A . P . ( i i ) b + c - c a , c + a - b b a + b - c c a r e i n A . P . p r o v e t h a t 1 a , 1 b , 1 c a r e i n A . P .