If first term ,second , and last term of an A.P be a ,b , c, respectively . Then show that the sum is 

(b+c-2a) (a+c) / 2(b-a) .

Q. If first term ,second , and last term of an A.P be a ,b , c, respectively . Then show that the sum is 

 [ (a+c) * (b+c - 2a) ]  /  2 *(b-a) .

 
ANS :   AP : a , b ............ ,c
 First term (a1) = a
 Common Difference (d) = (b - a)
 Last term ie, nth term (an) = c
Apply the Formula an = a1 + (n-1)*d
 
So :  c a + (n-1) *[b-a]
 
   c - a = (n-1) *[b-a]

  (n-1) =  (c-a) / [b-a]

=> n = { (c-a) / [b-a]  } + 1  = { (c-a) + (b-a) }  / [b-a]  =  {b+c - 2a} / [b-a] 

Therefore : n = {b+c - 2a} / [b-a]  -------(1)

To determine Sum of n term (Sn)   => Apply : Sn  =  [n / 2 ] * { a1 + an }

 Sn  =  [n / 2 ] * { ac } -----------(2)

From (1) : n = {b+c - 2a} / [b-a]  , Substitute  in (2) .

 Sn  =  [ {n} / 2 ] * { ac }

 Sn  =  [  {b+c - 2a} / [b-a]  } * (1 / 2) ] * { ac }

Sn = [ (a+c) * (b+c - 2a) ]  /  2 *(b-a)

Hence Sum of n terms  : Sn  = [ (a+c) * (b+c - 2a) ]  /  2 *(b-a)

(HENCE PROVED) 

IF YOU ARE SATISFIED DO GIVE THUMBS UP.

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thaxxxxxxxximad 93     i have also seen your alternate methos in sol. of n c e r t  can u plz  try my more questions

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