Register Now

Lorem ipsum dolor sit amet, consectetur adipiscing elit. Morbi adipiscing gravdio, sit amet suscipit risus ultrices eu. Fusce viverra neque at purus laoreet consequa. Vivamus vulputate posuere nisl quis consequat.

If first term of an A.P. is a, second term is b and last term is c, then show that sum of all terms is  ${(a-c)(b+c-2a)}\over 2(b-a)$

Print or Save

Sol.    For the given A. P. a, b, ......., c
First term = a
Common difference d = b – a.
Last term tn = c
∴          tn = a + (n –1)d
c = a + (n –1) (b – a)
c – a = (n –1) (b – a)

n – 1 = $(c – a)\over(b-a)$

n =  $(c – a)\over(b-a)$+1

n =  $c – a + b – a\over(b-a)$       [Taking LCM)

n =  $b + c – 2a\over{(b-a)}$            ...(I)

Sn =  ${n\over2}[t1+tn]$   [Formula]

From (I) and t1 = a, tn = c

Sn =  ${{b + c – 2a}\over2(b – a)}[a+c]$

Sn =   $(b + c – 2a) (a + c)\over2(b – a)$   [From (I)]

∴       Sn =   $(a + c)(b + c – 2a)\over2(b-a)$

∴        Hence proved

Arithmetic Progression September 28 , 2018 0 Comments 372 views