Question

The first, second and last term of an A.P are a,b,c respectively. Then prove that the sum is (a+c) (b+c-2a) /2(b-a).

Answer 1

★ ARITHMETIC PROGRESSION ★

Standard formula for determining the sums of nth numbers in A.P.

S(n) = n/2 [ 2a + ( n - 1 )d ]

First term : a
second term : b
LAST term : c

Common difference : b - a

T( n ) = a + (n - 1) d

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

HENCE , n = c-a/ b-a + 1

Now , obtaining the sum :

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

will further result in :

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

HENCE PROVED

★✩★✩★✩★✩★✩★✩★✩★✩★✩★✩★

Answer 2

Hey!!

first term = a,
second term = b
last term = c
Common difference d= (b-a)

an = a + (n-1) d

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

=> c = a + nb -an -b +a

=> c = 2a -an +nb -b

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

Sn = n/2 (a+ l)

=> Sn = (b+c-2a) (a+c) / 2(b-a)

hope it helps!!