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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) - Mathematics

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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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अध्याय 5: Arithematic Progressions - Exercise 5.4 [पृष्ठ ५७]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 10
अध्याय 5 Arithematic Progressions
Exercise 5.4 | Q 7 | पृष्ठ ५७

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