Question

Show that sum of an AP whose 1st term is a, the 2nd term b and the last term c, is equal to
$\frac{(a + c)(b + c - 2a)}{2(b - a)}$
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Answer 1

Answer:

first term = a

second term = b

last term = c

so, common difference = (b-a)

now, let last term be the N

term of the AP L = a + (n-1) d

c = a + (n-1) (b-a)

c = a + nb -an -b +a

c = 2a -an +nb -b

c = 2a -b +nb -an

n = (b+c-2a) / (b-a)

now, sum of n terms = n/2 (first term + last term)

Sn = (b+c-2a) (a+c) / 2(b-a)