Sequence and series Example
Question 1: If 4,7,10,13,16,19,22……is a
sequence, Find:
Ⓐ Common difference
Ⓑnth term
Ⓒ21st term
Solution: Given sequence is, 4,7,10,13,16,19,22……
Ⓐ) The common difference
d=A2-A1
= 7 – 4
= 3
Ⓑ) The nth term of the arithmetic sequence is denoted by the term An and is given by An = A₁ + (n-1)d, where “A” is the first term and d is the common difference.
An = 4 + (n – 1)3
= 4 + 3n – 3
= 3n + 1
c) 21st term as:
An = A1 + (n-1)d
A21 = 4 + (21-1)3
= 4+60
= 64
Question 2: Consider the sequence 1, 4, 16, 64, 256, 1024….. Find the common ratio and 9th term.
Solution:
The common ratio
r=G2/G1
r= 4/1
= 4
The preceding term is multiplied by 4 to obtain the next term.
The nth term of the geometric sequence is denoted by the term Gn and is given by Gn = r^-¹G1
where a is the first term and r is the common ratio.
Here G1= 1, r = 4 and n = 9
So, 9th term is can be calculated as
Gn = r^-¹G1
G9 = (4)⁹-¹(1)
= 4⁸
= 65536
Learning about mathematical concepts can be so much fun😁😁.
3. The sum of three numbers which
are consecutive terms of an A.P. is
21. If the second number is reduced
by 1 while the third is increased by 1,
three consecutive terms of a G.P.
result. The three numbers is
Ⓐ) 3,7, 11 Ⓒ)1,7,11
Ⓑ) 12, 7, 2 Ⓓ) a and b
solution:
Let the three numbers in A.P. is
A-B, A, A+B
B=reduce 1 or increase 1and A is require number.
A-B, A, A+B
Given፦ (A-B)+A+(A+B) = 21
3a =21
a= 21÷3
=7
The numbers are፦7-B, 7,7+B
If the second number is reduced by 1 while the third number is increased by 1, the resulting numbers are 7- B, 6, 8+B
which are given to be in G.P
6/7-B=8+B/6
36= (7-B) (8 + B)
36= 56-B+B²
B²-B+20
(B-5) (B+4)=0
B=-5, B=4
When B=-5;
7-B, 7,7+B
7-(-5),7,7+(-5)
7+5,7,7-5
12,7,2
The numbers are 12, 7, 2 and
When B= 4,
7-B, 7,7+B
7-4,7,7+4
3,7,4
When d= 4, the numbers are 3, 7, 11.
Answer፦d
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