9 Opt Maths: Sequence And Series

1. If S_n = 8n² + 16n + 12, find the value of t₂ + t₄+ t₆.
Solution:
Given ,
      S_n = 8n² +16n+12
We know, 
t_n =  s_n - s_n -1
    = 8n² +16n+12 - [ 8(n-1) ² +16 (n-1) + 12]
    = 8n² +16n+12 - [ 8n ² +16n +8 + 16n-16+12]
    = 8n² +16n+12 - 8n ² - 4
    = 16n + 8 .
Now,
    t₂ = 16 × 2+8 = 40   (put n = 2)
    t₄ = 16 × 4+8 = 72   (put n = 4)
    t₆ = 16 × 6+8 = 104   (put n = 6).
∴  t₂ + t₄ + t₆ = 40+72+104 =216.

2. The n^th term of a sequence is (-1)n-1.15nn2-4.. Find its third and eighth terms.
Solution:
Here, the n^th term of a sequence tn=(-1)n-1.15nn2-4
Now, the 3^rd term of a sequence t3=(-1)3-1.15×332-4      [ Since, put  n = 3.]
                                                           =(-1)2.459-4
                                                           =1×455
                                                           =9.
The 8^th term of a sequence t8=(-1)8-1.15×882-4     
[ Since, put n = 8.]
                                                   =(-1)7.12064-4
                                                   =-1×12060
                                                   =-2.

3. The n^th term of a sequence is (-1)n-1.75nn2-n-5. Find its 5th and 6th terms.
Solution:
Here, the n^th term of a sequence tn=(-1)n-1.75nn2-n-5
Now, the 5^th term of a sequence t5=(-1)5-1.75.552-5-5 
                                                           =(-1)4.75.525-10
                                                           =1×75×515
                                                            =25.

The 6^th term of a sequence t6=(-1)6-1.75×662-6-5
                                                   =(-1)5.75×636-11
                                                   =-1×75×625
                                                   =-18.

4. u1, u2, u3,…. are the terms of sequence where un+1=1-1un u1=3 find the values of u2 and u3.
Solution:
We have, u1=3
Here, un+1=1-1un……………i.

For n=1,

u1+1=1-1u1
⇒ u2 =1

-13= 3-13=23

For n = 2,

u2+1=1-1u2
⇒ u3=1-123=1-32=2-32=-12
Hence, the required value of u₂ and u₃ are 2/3 and -1/2.

5. Find the u2 and u3 of a sequence if its nth term is defined as un=3un-1-1 and u1=2.
Solution: 
Given,
u₁=2 
&   un =3u_(n-1)-1
For n = 2
u₂=3u_(2-1)-1 

      =3u₁-1=3×2-1=5 

For n=3
u₃=3u_(3-1)-1 
     =3u_2-1=3×5-1=14 .
 Thus, u2 = 5 and  u3 = 14

6. Find the nthterm and 12thterm of the sequence
-1,2,-3,4,…
Solution:
Here, the given sequence is-1, 2, -3 , 4,….
1stterm=t1=-1=(-1)1.1
2ndterm=t2=2=(-1)2.2
3rdterm=t3= -3=(-1)3.3
Therefore, nthterm=tn=(-1)n.n
Now, 12thterm=t12=(-1)12.12=12

7. Write down the next two terms of the sequence 2,5, 10, 17,
Solution: 
Here, the given sequence is 2, 5, 10, 17,
We have, t1=2
t2=5=2+3=t1+3
t3=10=5+5=t2+5
t4=17=10+7=t3+7
Now,
t5=t4+9=17+9=26
t6=t5+11=26+11=37

8. Denote the given series 2+4+8+16+32+64 by Σ notation.
Solution:
Here, the given series=2+4+8+16+32+64
We have, first term=t1=2=21
Second term=t2=4=22
Third term=t3=8=23
Hence, nthterm=2n
In the given series, number of term=6
Hence, the given series can be written in form of summation as:
6    
∑2^n
n=1

9. If S_n = 4n² + 5n + 8, find the value of t³ + t⁵ + t⁷.
Solution:
Given S_n = 4n² + 5n + 8
Now, 
t_n = S_n - S_n-1 
= (4n² + 5n + 8) - [4(n - 1)² + 5(n - 1) + 8]
= 4n² + 5n + 8 – [4n² - 8n + 4 + 5n – 5 + 8]
= 8n + 1
Again, t₃ = 8 × 3 + 1 = 25
            t₅   = 8 × 5 + 1 = 41
                  t₇   = 8 × 7 + 1 = 57
 ∴ t₃ + t₅  +  t₇    =  25 + 41 + 57
                                = 123 . 
 
10. If S_n = 8n² + 16n + 12, find the value of t₂ + t₄+ t₆.
Solution: 
Given,
 S_n = 8n² + 16n +12
Now,
  t_n = S_n - S_n - 1
       = 8n² + 16n +12 - [8(n - 1)² + 16(n - 1) + 12]
      = 8n² + 16n + 12 – (8n² - 16n + 8 + 16n - 16 + 12)
       = 8n² + 16n + 12 - 8n² - 4
        = 16n + 8
 ∴ t₂ = 16 × 2 + 8 = 40
    t₄ = 16 × 4 + 8 = 72
   t₆ = 16 × 6 + 8 = 104
 Thus,  t₂ + t₄ + t₆ = 40 + 72 + 104 = 216 

11. If in a seqⁿ t₆ =50 and t_n=2t_{n - 1} + 5n² + n then find the first five terms of the sequence.
 Solution: 
Given, t₆ = 50  and
            t_n = 2t_{n - 1} + 5n² + n, 
          For n = 6, 
t₆   = 2 t_6-1 + 5 × 5² + 5 
⇒ 50 = 2 t₅  +  130
⇒ 2 t₅  = - 80
⇒ t₅  = - 40
Again, for n = 5,
t₅   = 2 t₄ + 5 × 4² + 4 
⇒ - 40 = 2 × t₄  +  100 + 4
⇒ 2 t₄  = 144
⇒ t₄  = 72
Again, for n = 4,
t₄ = 2 t₃ + 2 × 42 + 4
⇒  72 = 2 t₃ + 36
⇒ t₃ = 18
Again, for n = 3
t₃ = 2 t₂ + 5 × 32 + 3
⇒  18 = 2 t₃ + 48
⇒ 2t₂ = - 30
t₂ = -15
Again, for n = 2
t₂ = 2 t₁ + 5 ×22 + 2
⇒  -15 = 2 t₁ + 22
⇒ t₁ = - 372
∴1st five term are
- 372 , -15 , 18, 72, & -40.

12. In a sequence if t₈ = 24 and t_n = 2t_n+2 + 5n² + n. Find the first three even terms of the sequence.
Solution:
Given, t₈ = 24.
t_n = 2t_n+2 + 5n² + n, 
For n = 6, 
t₆    = 2 t₈ + 5 × 6² + 6 
       = 2× 24 + 5 × 36 + 6
       = 234.
For, n = 4, 
t₄    = 2 t₆ + 5 × 4² + 4 
 t₄   = 2 × 234 + 80 + 4
      = 552.
For, n = 2, 
t₂   = 2 t₄ + 5 × 2² + 2
      = 2 × 552 + 20 + 2
      = 1126.
 ∴ 1^st three even term are 1126, 552 and 234.
 
13. Study the given pattern of the given figures:
a.    Add one more figures in the same pattern. 
b.    Find the formula of the nth term of the sequence. 
c.    Find the 10th term of the same sequence. 
 Solution:
Given,
t₁ = 1
t₂ = 4   
t₃ = 7
t₄ = 10
a)

b)  Here, 
         a = t₁ = 1
        d = t₂ - t₁ = 4 - 1 = 3
   ∴ t_n = a + (n - 1)d
          = 1+ (n - 1) × 3
          = 1+ 3n -  3
          = 3n - 2
 c).      t₁₀ = 3 × 10 - 2
             = 28.


14. In the given pattern of triangular numbers:
a. Add one more figure in the same pattern that comes immediately to the next term in the sequence. 
b. Find the formula for the nth term of sequence. 
Solution:
Given 
t₁ = 1 = a
t₂ = 3
t₃ = 6
t₄ = 10

Do yourself and find  

n
2
(n+1)

15. In the following pattern of numbers, 
a. Add one more figure in the same pattern. 
b. Find the formula for the nth term of the sequence. 
 Solution:
Given 
   t₁ = a =  1
   t₂ = 5
   t₃ = 9
   t₄ = 13.

a)

b)
Here, 
     t₁ = a = 1
     t₂ = 5
    d = t₁ - t₂ = 5 - 1 = 4 

  ∴  t_n = a + (n - 1)d
         = 1 + (n - 1) 4
        = 4n -  3.

16. Study the following patterns of numbers:
a. Add two more patterns in the sequence. 
b. Find the formula for the nth dots in the term of the sequence. 
Solution: 

a)

b)
Here,
     t₁ = a =3
     t₂ = 5
    ∴  Common difference (d) = t₂ - t₁
                                               = 5 - 3
                                               = 2.

 General term (t_n)  = a + (n - 1)d
                               = 3 + (n - 1) 2
                               = 3 + 2n -  2
                               = 2n +  1.
 
17. Study the following patterns:
a. Add two more patterns in the sequence. 
b. Find the formula for the nth dots in the term of the sequence. 
c. Find the 10th term of the sequence. 
 Solution: 
a) Next two terms are, 

b)
t₁ = 1 = 12 (1 + 1)
t₂ = 3 = 22 (2 + 1)
t₃ = 6 = 32 (3 + 1)
t₄ = 10 = 42(4 + 1)

Replace 4 = n we get,
t_n = n2 (n + 1)

Now,
10th term (t₁₀) = 102 (10 + 1)
                        = 5 × 11
                        = 55 .


Patterns will be uploaded soon.