Question

25. Consider the numbers between 100 and 500 which when divided by 7 gives a remainder 3.
a) What is the first number of the sequence?
b) What is the last number of the sequence?
c) How many terms are there in this sequence?
d) Find the sum of these terms?​

Answer 1

Step-by-step explanation:

a. 105

b.497

Answer 2

Solution:

105, 112, 119, 126, 133, 140, 147, 154, 161, 168, 175, 182, 189, 196, 203, 210, 217, 224, 231, 238, 245, 252, 259, 266, 273, 280, 287, 294, 301, 308, 315, 322, 329, 336, 343, 350, 357, 364, 371, 378, 385, 392, 399, 406, 413, 420, 427, 434, 441, 448, 455, 462, 469, 476, 483, 490, 497,

A. 105

B. 497

C. Tn = a + (n - 1)d
First term, a = 105
Common difference, d = 7
497 = 105 + (n - 1)7
497 = 105 + 7n - 7
497 = 98 + 7n
497 - 98 = 7n
399 = 7n
n = 399/7
n = 57

D. Sn = n/2 [2a + (n - 1)d] or
Sn = n/2 (a + l)
Sn = 57/2 ( 105 + 497)
Sn = 57/2 (602)
Sn = 28.5 (602) = 17157

Hope this will be helpful.