Question

2. Show that the sum of an AP whose first term is 'a', the second term is 'b' and the last term is equal to "c" is equal to
to ( a + c)(b+ c-2a)
2(b-a)​

Answer 1

Answer:

Explanation:

a,b....c

First term=a

Common diff.=b-a

Tn=a+(n-1)d

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

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

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

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

Taking L.C.M.

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

Therefore,

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

Sn=n/2(a+l) where l=last term

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

Hence, proved

Answer 2

plz refer to this attachment