# 3 questions

0 votes
Given t1 = -23, n = 25, and tn = 57, find the sum of the arithmetic series.
 A. 125 B. 250 C. 325 D. 425

Find a formula for tn so that the first four terms are 15, 25, 35, 45.
 A. tn = 15n B. tn = 10n - 5 C. tn = 10n + 5 D. tn = 15n - 5
Find the first four terms of the given sequence tn = (-1)n(n).
 A. -1, -2, -3, -4 B. -1, 2, -3, 4 C. 1, ½, 2/3, ¾ D. 1, 2, 3, 4

asked Sep 11, 2014

## 3 Answers

+1 vote

Best answer

1)

Given t1 = -23, n = 25, and tn = 57.

Find the common difference by using the formula:

tn = t1 + (n-1)d

Substitute -23 for t1 , 57 for tn , and 25 for n.

57 = -23 + (25-1) d

24d = 80

d = 80/24 =10/3

Find the sum of the n terms.

Sn = (n/2)(2t1 + (n-1)d )

= (25/2)(2(-23) + (25-1)(10/3) )

= (25/2)(2(-23) + 80 )

= (25/2)(34)

= 425

So Option D is correct

answered Sep 11, 2014
selected Sep 12, 2014 by tonymate
+1 vote

2)

Given 15, 25, 35, 45.

Find the common difference by using the formula:

d =t2 - t1

Substitute 15 for t1 and 25 for t2.

d = 25 - 15 =10

Find the nth term by using the formula.

tnt1 + (n-1)d

Substitute 15 for t1 and 10 for d .

tn = 15 + (n-1)(10)

= 15 + 10n - 10

= 5 + 10n

So Option C is correct

answered Sep 11, 2014
+1 vote

The given sequence tn = (-1)n(n).

Find the first term of the sequence.

Substitute 1 for n in sequence.

t1 = (-1)1(1) = -1

Find the second term of the sequence.

Substitute 2 for n in sequence.

t2 = (-1)2(2) = 2

Find the third term of the sequence.

Substitute 3 for n in sequence.

t3 = (-1)3(3) = -3

Find the fourth term of the sequence.

Substitute 4 for n in sequence.

t4 = (-1)4(4) = 4

So sequence is -1,2,-3, and 4

So Option B is correct.

answered Sep 11, 2014