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

Sum

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))

Solution

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,

S_n = 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.

Concept: Sum of First n Terms of an A.P.
Is there an error in this question or solution?

APPEARS IN

NCERT Mathematics Exemplar Class 10
Chapter 5 Arithematic Progressions
Exercise 5.4 | Q 7 | Page 57
Share