Question

if a,b,c be the first ,third and nth terms respectively of an AP prove that the sum to n terms is (c+a)/2+(c+a)(c-a)/b-a

Answer 1

Answer:

Given that a = first term

and

b = T3 = a + 2d

and

c = Tn = a + ( n - 1)d.

Now, T3 - T1 = b - a

⇒ 2d = b - a.

d = ( b -a) / 2.

Substitute d in nth term, we get

c = a + ( n - 1) d

⇒ (c - a) = (n - 1) ( b -a) / 2.

⇒ 2(c - a) / ( b -a) = (n - 1)

∴ n = ( b + 2c - 3a) / ( b - a).

Sum of n terms

Sn = ( n / 2) [ a + l ]

⇒ Sn = ( ( b + 2c - 3a) / 2( b - a).) [ a + c ] [ substitute n and l ]

⇒ Sn = ( b + 2c - 3a)( a + c ) / 2( b - a).

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

Sn = [(c + a) / 2 + (c2- a2) / (b-a)].

Hence proved.