Question

The sum of r terms of the series ( a -1 )+(a-2)................ is

Answer 1

Hey there!

(a-1) , ( a-2) , (a-3),... are in Arithmetic progession.

Now, Common difference = (a-2)-(a-1) = a-2-a+1 = -1

We know that, Given a as the first term and d the common difference, Sum of n terms of an A. P = n/2 [ 2a + ( n-1)d ]

Now,

Sum of " r " terms of the series
= r/2 [ 2(a-1)+(r-1)(-1)
= r/2 [ 2a - 2 -r +1 ]
= r/2 [ 2a -r -1 ]
=[ 2ar-r²-r]/2
Hope helped!

Answer 2

(a-1), (a-2), (a-3).........

First term = a-1

Common difference = T2 - T1

= (a - 2) - (a-1)

= a - 2 - a + 1

= - 1

Now,

Sr = r/2 [ 2×First term + (r-1) d]

=> Sr = r/2 [ 2 × (a-1) + (r-1) (-1)]

=> Sr = r/2 [ 2a - 2 - r + 1]

=> Sr = r/2 [ 2a - r - 1]

=> Sr = (2ar - r^2 - r) / 2