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A succession of numbers each of which is formed according to the same fixed law is called a sequence. In sequence, there is a simple rule for going from one number to another.
Let us consider the following set of numbers,
In (1) each number is greater than the preceding and by 2. Each number is formed by adding 2 to the preceding term. So, the number next to 7 is 7+ 2 = 9.
In (2) each number is double the preceding number. Each number is formed by multiplying the preceding number by 2. So, the number next to 16 is 16x2= 32.
But the set of numbers in (3) does not follow any rule. So, we cannot guess the number next to 17. Here, the set of numbers in (a) and (b) are arranged in order by some rule. So, these are the examples of sequences of numbers. But the set of numbers in (c) does not form a sequence.
Thus, an ordered list of the numbers connected by a rule is called a sequence of the numbers. This is the traditional definition of the sequence.
Generally, there are tow types of sequence. They are given below:
Each successive pair of terms having a common difference in a sequence is called a linear sequence.
A quadratic sequence is a sequence of number in which there is same difference in the number.
Here,
Let the given sequences be a_{1}, a_{2}, a_{3}, a_{4}............
Now, a_{1}=5
a_{2}= 2=5-3
a_{3}= -1=2-3
a_{4}=-4=-1-3
Similarly the next term is a_{5}= -4 -3 = -7.
Again from the given pattern
a_{1}=5= 8-3 × 1
a_{2}=2 =8-3× 2
a_{3}= -1=8 - 3× 3
a_{4} = -4 =8-3× 4
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⋅ ⋅ ⋅
⋅ ⋅ ⋅
a_{n}=8 - 3×n
∴ n^{th} term (a_{n}) = 8 - 3n.
Now 12^{th} term (a_{12}) = 8 - 3× 12 = 8 - 36 = -28.Ans
Here,
Let the given sequences be a_{1}, a_{2}, a_{3}, a_{4}............
Now, a_{1} = 2
a_{2} = 4 =2 + 2
a_{3} = 6 = 4 + 2
a_{4} = 8 = 6 + 2
Similarly the next term is a_{s} = 8 + 2 =10.
Again from the given pattern.
a_{1}=2=2 × 1
a_{2} = 4 = 2× 2
a_{3}= 6 = 2× 3
a_{4} = 8 = 2× 4
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a_{n}=2n
∴ n^{th} term a_{n}= 2n
Again, 12^{th} term a_{12} = 2 × 12
= 24 Ans.
soln: let the sequences be a_{1},a_{2},a_{3},a_{4},a_{6}..........
n^{th }term (a_{n})=4n-3
∴ a_{1}=4×1-3=4-3=1
a_{2}=4×2-3=8-3=5
a_{3=}4×3-3=12-3=8
a_{4}=4×4-3=16-3=13
a_{5=}4×5-3=20-3=17
a_{6}=4×6-3=24-3=21
∴ The first 6 terms of the sequences are : 1,5,9,13,17 and 21 Ans.
soln: Let the sequences be a_{1},a_{2,}a_{3,...........}
n^{th }term (a_{n}) = n^{2}+2
∴ a_{1}=1^{2}+2=1+2=3
a_{2}=2^{2}+2=4+2=6
a_{3}=3^{2}+2=9+2=11
a_{4}=4^{2}+2=16+2=18
a_{5}=5^{2}+2=25+2=27
a_{6}=6^{2}+2=36+2=38
∴ The first 6 terms of the sequences are:3,6,11,18,27 and 38 Ans.
Here
let the sequence be a_{1},a_{2},a_{3}........
n^{th}term (a_{n})=(-2)^{n+1}
a_{1} =(-2)^{1+1} =(-2)^{2 }=4
a_{2}=(-2)^{2+1}=(-2)^{3}=-8
a_{3}=(-2)^{3}^{+1}=(-2)^{4}=16
a_{4}=(-2)^{4}^{+1}=(-2)^{5}=-32
a_{5}=(-2)5^{+1}=(-2)^{6}=64
a_{6}=(-2)^{6}^{+1}=(-2)^{7}=-128
∴ The first 6^{th} term of the sequences are;
4,-8,16,-32,64 and -128 Ans.
Here given sequences,
a_{n}=2n
5 terms of the sequences by putting n =1,2,3,4, and 5 we get,
a_{1}=2×1=2, a_{2}=2 ×2 =4, a_{3}=2×3=6,
a_{4} = 2 × 4 =8 and a_{5} =2×5=10
Sum of the 5 terms.
S_{5}=a_{1}+a_{2}+a_{3+}a_{4+}a_{5}
=2 + 4 + 6 + 8 + 10
=30 Ans.
Here, given
n^{th} term of the sequences a_{n}=(-1)^{n} 2^{n}
Putting n=1,2,3,4 and5 we get,
a_{1}=(-1)^{1}.2^{1}=-2
a_{2}=(-1)^{2}.2^{2}=4
a_{3}=(-1)^{3}.2^{3}=-8
a_{4}=(-1)^{4}.2^{4}=16
a_{5}=(-1)^{5}.2^{5}=-32
Sum of the 5 terms
S_{5}= a_{1}+a_{2}+a_{3}+a_{4}+a_{5}
= -2+4+-8+16-32
= -42+20
=-22. Ans.
Here, given
n^{th} term of the sequences a_{n}=(-1)^{n+1} 2^{n}
Putting n=1,2,3,4 and5 we get,
a_{1}=(-1)^{1+1}.2^{1}=(-1)^{2}.2=2
a_{2}=(-1)^{2+1}.2^{2}=(-1)^{3} 4=-4
a_{3}=(-1)^{3+1}.2^{3}=(-1)^{4}.8=8
a_{4}=(-1)^{4+1}.2^{4}=(-1)^{5}.16=-16
a_{5}=(-1)^{5+1}.2^{5}=(-1)^{6}.32=32
Sum of the 5 terms
S_{5}= a_{1}+a_{2}+a_{3}+a_{4}+a_{5}
= 2-4+-8-16+32
= 42-20
=22. Ans.
Here,
Let the sequences be a_{1,}a_{2},a_{3...........}
n^{th} term (a_{n}) =2n-1
∴ a_{1} = 2 × 1 -1 = 2 - 1= 1
a_{1} = 2 × 2 -1 = 4 - 1= 3
a_{2} = 2 × 3 -1 = 6 - 1= 5
a_{3} = 2 × 4 -1 = 8 - 1= 7
a_{4} = 2 × 5 -1 = 10 - 1= 9
a_{5} = 2 × 6 -1 = 12 - 1= 11
∴The first 6 terms of the sequences are:1,3,5,7,9 and 11. Ans.
Find the U_{2} and U_{3} of a sequence if its n^{th} term is defined as U_{n}= 3U_{n-1}-1 and U_{1}=2
5,14
10,2
4,12
5,12
Find the sum of first four terms (S_{4}):
t_{n}=4n+5
33
60
57
20
Find the sum of first four terms (S_{4}):
t_{n}=3n+2
78
38
22
12
Find the sum of first four terms (S_{4}):
t_{n}=5n+6
56
89
74
23
Find the sum of first four terms (S_{4}):
t_{n}=6n-3
48
14
39
84
Find the sum of first four terms (S_{4}):
t_{n}=2^{n}
30
23
46
55
Find the sum of first four terms (S_{4}):
t_{n}=n^{2}
20
64
30
21
Find the sum of first four terms (S_{4}):
t_{n}=n^{2}+4
36
56
46
26
Find the sum of first four terms (S_{4}):
t_{n}=n^{2}-2
38
9
11
22
Find the sum of first four terms (S_{4}):
t_{n}=(frac{2n}{n+1})
8(frac{13}{20})
6(frac{13}{20})
5(frac{9}{22})
5(frac{13}{20})
Find the sum of first four terms (S_{4}):
t_{n}=(frac{2n-1}{2n+1})
2(frac{564}{315})
2(frac{134}{315})
2(frac{134}{115})
3(frac{194}{515})
Find the n^{th} term the 3, 6, 9, 12, . . . . . . . . sequences
FInd the n^{th} term of the quadratic sequences 1, 7, 17, 31, 49, . . . . . . . . . .
Find the frist term of the sequences whose n^{th } term is t_{n} = n - 1
ASK ANY QUESTION ON Sequences
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