A day full of math games & activities. Find one near you.

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)]

[(a+c)(b+c-2a)]/2(b-a)

Solution:

Given, the AP is a, b, c,.....

We have to prove that the sum of the AP is equal to [(a+c)(b+c-2a)/2(b-a)]

The nth term of the series in AP is given by

aₙ = a + (n - 1)d

Here, first term, a = a

Last term, l = c

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

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

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

n = [(c - a)/(b - a)] + 1

If l is the last term of an AP, then the sum of the terms is given by

S = n/2[a+l]

So, S = ([(c - a)/(b - a)] + 1)/2[a + c]

= ((c - a)+(b - a)[a + c])/2(b - a)

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

S = [(a+c)(b+c-2a)]/2(b-a)

Hence proved.

✦ Try This: Find the sum of an AP whose first term is d, the second term e and the last term f

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 5

NCERT Exemplar Class 10 Maths Exercise 5.4 Problem 7

Summary:

It is proved 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)]

☛ Related Questions: