First second and last term of Ap are a, b,2a respectively, find its sum

  (3ab)/(2(b-a))

Step-by-step explanation:

The n-th term of an arithmetic progression is ...

  an = a1 +d(n -1)

Then the value of n is ...

  n = (an -a1)/d +1

The sum of an arithmetic progression is the product of the number of terms and the average of the first and last terms. In this sequence, the common difference d is ...

  d = (b -a)

So, the sum is ...

  Sn = (a +2a)/2·((2a -a)/(b -a) +1)

  Sn = (3ab)/(2(b-a)) . . . . sum of the arithmetic progression

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Example:

The sequence 1, 1.5, 2 has ...

  a = 1, b = 1.5

Its sum is given by the above formula as ...

  Sn = 3(1)(1.5)/(2(1.5 -1)) = 4.5/(2(.5)) = 4.5 = 1 + 1.5 + 2 . . . . yes

idk

step-by-step explanation:

c > -10.55

step-by-step explanation: