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Q) The first , second and middle term of an AP are a,b,c respectively. Sum of all terms is

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

      C) 2c(b-a)/c-a.               D) 2b(c-a)/b-a

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by (15 points)
B)

Let's Think , The Common Difference of this AP Series = d

Now , first term = a

           second term = b

So , b = a + d

  Or ,     d = b-a

Now , if there are total ( 2n +1 ) th term in the series , the middle term will be ( n+1) term .

Therefore , c = a + nd = a + n(b-a)

        From this we can get , n = (c-a) / (b-a)

So , the total sum = [(2n+1)/2][2a + 2nd ]

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

You can find out the last term by putting the value of n and d in the above term and rearranging that .

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