Consider the arithmetic sequence : 184, 178, 172, ...a) Write the algebraic form? b) How many positive numbers are in the sequence?
Answers
Answer:
Given arithmetic sequence is 9, 15, 21,.
First term = a = 9
Common difference = d =15 9 = 6
(A) For natural number n
nth term = [(n−1)x6]+9=6n+3 where n=1,2,3,
Hence algebraic form of the sequence 9,15,21,. is x
n
=6n+3.
(B) nth term =6n+3
25
th
term =6(25)+3=150+3=153
(C) Sum of first n terms is given by,
S
n
=
2
1
n[X
1
+X
n
]
Sum of 24 terms is
S
24
=
2
1
n[X
1
+X
2
4]=
2
25
[9+147]=1950
Sum of 50 terms is
S
50
=
2
1
n[X
1
+X
50
]=
2
50
[9+6(50)+3]=7800
Sum of terms from twenty fifth to fiftieth
=S
50
−S
24
=7800−1950
=5850
(D) Let sum of first n terms is 2015,
S
n
=
2
1
n[X
1
+X
n
]
2015=
2
1
n[9+6n+3]
2015=3n
2
+6n
3n
2
+6n−2015=0
Solving above equation for n, the value of n is not a natural number.
Given arithmetic sequence 184, 178, 172....
Here, the difference between the numbers in the sequence is 6.
(a) The algebraic form is a, a - 6, a - 2(6)...
(b) The positive numbers in this sequence are 30.
184, 178, 172,.... 16, 10, 4,... -2, -8