the first second and last term of an ap are a b c respectively show that the sum of the ap is (b+c-2a)(a+c)/2(b-a)
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31
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)]
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