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

User Wil Tan
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1 Answer

6 votes

Answer:

(3ab)/(2(b-a))

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