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Question
If first term of an A.P. is a, second term is b and last term is c, then show that sum of all terms is \[\frac{\left( a + c \right) \left( b + c - 2a \right)}{2\left( b - a \right)}\].
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Solution
a, b, ..., c
t1 = a, d = b - a, tn = c
We know that
tn = a + (n - 1)d
c = a + (n - 1)(b - a)
`(c-a)/(b-a)=n-1`
`(c-a)/(b-a)+1/1=n`
`(c-a+b-a)/(b-a)=n`
∴ n = `(c+b-2a)/(b-a)` ...(1)
Now,
Sn = `n/2[t_1 + t_n]`
Sn = `(c+b-2a)/((b-a)2)[a+c]`
Sn = `((a+c)(b+c-2a))/(2(b-a))`
Hence proved.
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