Question

17.Show that the sum of an arithmetic series whose first term is a, second term b and the last term is c is equal to

Answer 1

Steps :
1) Series :a, b, ........c.
Let ' n' be total number of terms.
First term, a(1) = a
Second term,a(2) = b
Last term, a(n) = c
Common difference,
d = a(2)-a(1)
= b-c

2) We know that,
a(n) = a + (n-1)d
=> c = a + (n-1) (b-c)
=> (n-1) = (c-a) /(b-a)
=> n = (c+b-2a)/(b-a)

3) Now,
S(n) = n/2( a + a(n))
=$\frac{(b+c-2a)}{2(b-a)} (a+c)$

Hence Proved.

Answer 2

Solution :

It is given that , In A.P

First term = a ,

Second term = b ,

last term = c .

Common difference ( d ) = b - a

Let number terms in A.P = n

Last term = c

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

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

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

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

=> n = [( c - a + b - a )/( b - a ) ]

n = ( c + b - 2a )/( b - a ) ---( 1 )

Sum of n terms = Sn

Sn = n/2[ a + c ]

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

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

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