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first term = a
second term =b
last term = c
common difference = b-a
Now,
tn = a+(n-1)d
c= a+(n-1)(b-a)
or ,(c-a)= (n-1)(b-a)
or, (c-a)/(b-a) = n-1
or, n= (c-a)/(b-a) +1
or, n = (c+b-2a)/(b-a)......... (i)
we know,
sum of n terms = n/2 (first term+last term)
or sum of all terms = {(c+b-2a)/2 (b-a) }(a+c)
or, sum of all terms = (c+a)(c+b-2a)/2 (b-a)
hence proved .
second term =b
last term = c
common difference = b-a
Now,
tn = a+(n-1)d
c= a+(n-1)(b-a)
or ,(c-a)= (n-1)(b-a)
or, (c-a)/(b-a) = n-1
or, n= (c-a)/(b-a) +1
or, n = (c+b-2a)/(b-a)......... (i)
we know,
sum of n terms = n/2 (first term+last term)
or sum of all terms = {(c+b-2a)/2 (b-a) }(a+c)
or, sum of all terms = (c+a)(c+b-2a)/2 (b-a)
hence proved .
thakurayushi:
hey how we know the sum of all term is n/2
we have formula , In AP , Sn =n/2 (first term +last term )
If series is finite
ohh and how can u get 2b-a
at last
putting value of n from (i) in last formula
oh yaah i understand that
thank j
thank u
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