Hi there...Originally Posted byConfuzzled?

For Qn 1, for typing easeness, i will denote the Sigma sign n on the top, 1 underneath as (S1).

(S1) (2r-7)

= (S1) (2r) - (s1) (7)

= 2 (S1)(r) -7n

=2 (1/2)(n)(n+1) - 7n

=n2 + n - 7n

=n2 - 6n

=n(n-6)

For Qn 2, 1st term = a = 14 ; Sn = d = -6

Sn = n/2(2a + (n-1)d) => eq 1

Substitute the following values on top into eq 1 and solve for n.

For Qn 3,

1st Term = 2x + 1

2nd Term = 3x + 2

3rd Term = 5x - 1

As they are in arithmetic series, they will have a common difference.

Thus,

(3x + 2) - (2x + 1) = (5x - 1) - (3x + 2)

Solve the above equation for x.

Hope these can be of some help to u.