Question

7. 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)​

Answer 1

Answer:

Step-by-step explanation:first term =a

second term =b

last term =c

common difference=(b-a)

tn =a+(n-1) d

c=a+(n-1) d

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

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

( c+b-2a)/(b-a)=n              ---(1)

now ,

we know sum of n terms=n/2 (first term+last term)

put equation (1)value

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

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