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प्रश्न
Show that the sum of an AP whose first term is a, the second term b and the last term c, is equal to `((a + c)(b + c - 2a))/(2(b - a))`
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उत्तर
Given that, the AP is a, b, c
Here, first term = a,
Common difference = b – a
And last term, l = an = c
∵ an = l = a + (n – 1)d
⇒ c = a + (n – 1)(b – a)
⇒ (n – 1) = `(c - a)/(b - a)`
n = `(c - a)/(b - a) + 1`
⇒ n = `(c - a + b - a)/(b - a)`
= `(c + b - 2a)/(b - a)` ...(i)
∴ Sum of an AP,
Sn = `n/2 [2a + (n - 1)d]`
= `((b + c - 2a))/(2(b - a)) [2a + {(b + c - 2a)/(b - a) - 1}(b - a)]`
= `((b + c - 2a))/(2(b - a))[2a + (c - a)/(b - a) * (b - a)]`
= `((b + c - 2a))/(2(b - a)) (2a + c - a)`
= `((b + c - 2a))/(2(b - a)) * (a + c)`
Hence proved.
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