Question

11. The sum of progression (a+b), a, (a-b) n term
is
(a) [2a + (n-1)b]
(b) [2a + (3-n)b]
(c) [2a + (3-n)]
(d) [2a + (n-1)]
​

Answer 1

Step-by-step explanation:

Sn = n/2[2a+(n-1)d]

=n/2[2(a+b) + (n-1)-b]

=n/2[2a+2b-nb+b]

=n/2[2a+3b-nb]

=n/2[2a+(3-n)b]

First Term = a+b

D(difference)=a-(a+b)

=a-a-b

=-b

Answer 2

First Term = a+b

D(difference) = a - (a + b)

=a-a-b

=-b

so,

Sn = n / 2 * [2a + (n - 1) * d]

= n / 2 * [2(a + b) + (n - 1) - b]

= n / 2 * [2a + 2b - nb + b]

= n / 2 * [2a + 3b - nb]

= n / 2 * [2a + (3 - n) * b]

ans (b).