find the sum of the first n terms of the sequence. The sequences are either arithmetic or geometric.

Question

13. -1, 11, -121, ...; n = 9



14. 14, 8, 2,...; n=9

Answer 1

13.

The geometric sequence : -1,11, -121,.............,n = 9

The first term of the sequence : a = -1

The ratio r = -11

The number of terms in the arithmetic sequence : n = 9

The sum of the first n terms of the series = a(1 - r²) / (1 - r)

Substitute a = -1, r = -11 in the sum of n terms

                                                                          = -1(1 - (-11)ⁿ) / (1 - (-11))

                                                                          = -1(1 + 11ⁿ) / 12 ( if n is odd)

                                                                          = -1(1 - 11ⁿ) / 12 (if n is even).                                                                         

 

Comments

The sequence is - 1, 11, - 121, ... and n = 9.

The above sequence is geometric sequence, since r = a2 /a1 = a3 /a2 = - 11.

Answer 2

14.

The sequence of the arithmetic series :14,8,2,..........n = 9

The first term of the arithmetic series : a = 14

The common difference of series : d = 14 - 8 = -6

The number of terms in the series : n = 9

The sum of the arithmetic series : A = n / 2[2a + (n - 1)d

Substitute n = 9, a = 14, d = -6 in the above sum of the series

                                                               A = 9 / 2[2(14) + (9 - 1)(-6)

                                                                   = 9[ 14 - 24]

                                                                   = -90.

Comments

The common difference of series : d = a2 - a1 = 8 - 14 = - 6.