StudentAidUnit 6: Sequences and Series — Exercise 6 1
11th Class Mathematics · Unit 6: Sequences and Series
1.(i) Find the next four terms of the sequence: 12, 16, 20, … (ii) Find the next four terms of the sequence: 3, 1, −1, … (iii) Find the first three terms of the sequence with formula a_n = (n−3)(n+1) (iv) Find the first three terms given a_1 = −1 and a_{n+1} = \frac{3}{a_n + 2} (v) Find the first three terms given a_n = 8 - \frac{20}{3+n}
Given
(i) First term
a1 = 12
Common difference
d = 4
Second term
a2 = 12 + 4 = 16
Third term
a3 = 16 + 4 = 20
Fourth term
a4 = 20 + 4 = 24
Fifth term
a5 = 24 + 4 = 28
Sixth term
a6 = 28 + 4 = 32
Seventh term
a7 = 32 + 4 = 36
Result
Next four terms
24, 28, 32, 36
Given
(ii) First term
a1 = 3
Common difference
d = -2
a2 = 3 - 2 = 1
a3 = 1 - 2 = -1
a4 = -1 - 2 = -3
a5 = -3 - 2 = -5
a6 = -5 - 2 = -7
a7 = -7 - 2 = -9
Result
Next four terms
-3, -5, -7, -9
Formula
(iii) General term
an = (n-3)(n+1)
Working
For n=1
a1 = (1-3)(1+1) = (-2)(2) = -4
For n=2
a2 = (2-3)(2+1) = (-1)(3) = -3
For n=3
a3 = (3-3)(3+1) = (0)(4) = 0
Result
First three terms
-4, -3, 0
Given
(iv) First term
a1 = -1
Formula
Recurrence relation
an+1 = 3an + 2
Working
For n=1
a2 = 3a1 + 2 = 3-1 + 2 = 31 = 3
For n=2
a3 = 3a2 + 2 = 33 + 2 = 35
Result
First three terms
-1, 3, 35
Formula
(v) General term
an = 8 - 203+n
Working
For n=1
a1 = 8 - 203+1 = 8 - 204 = 8 - 5 = 3
For n=2
a2 = 8 - 203+2 = 8 - 205 = 8 - 4 = 4
For n=3
a3 = 8 - 203+3 = 8 - 206 = 8 - 103 = 24-103 = 143
Result
First three terms
3, 4, 143
2.(i) Write down the first three terms of the sequence a_n = 3n + 5 (ii) Write down the first three terms given a_{n+1} = 4a_n − 7 and a_1 = 3 (vi) Write down the first three terms given a_1 = 1 and a_{n+1} = (3a_n + 2)^2 (vii) Write down the first three terms given a_n = (−2n)^2 (viii) Write down the first three terms given a_n = (−1)^n 7n^2
Formula
(i) General term
an = 3n + 5
Working
For n=1
a1 = 3(1) + 5 = 8
For n=2
a2 = 3(2) + 5 = 11
For n=3
a3 = 3(3) + 5 = 14
Result
First three terms
8, 11, 14
Given
(ii) First term
a1 = 3
Formula
Recurrence relation
an+1 = 4an - 7
Working
For n=1
a2 = 4(3) - 7 = 12 - 7 = 5
For n=2
a3 = 4(5) - 7 = 20 - 7 = 13
Result
First three terms
3, 5, 13
Given
(vi) First term
a1 = 1
Formula
Recurrence relation
an+1 = (3an + 2)2
Working
For n=1
a2 = (3(1) + 2)2 = (5)2 = 25
For n=2
a3 = (3(25) + 2)2 = (77)2 = 5929
Result
First three terms
1, 25, 5929
Formula
(vii) General term
an = (-2n)2
Working
For n=1
a1 = (-2 × 1)2 = (-2)2 = 4
For n=2
a2 = (-2 × 2)2 = (-4)2 = 16
For n=3
a3 = (-2 × 3)2 = (-6)2 = 36
Result
First three terms
4, 16, 36
Formula
(viii) General term
an = (-1)n 7n2
Working
For n=1
a1 = (-1)1 × 7(1)2 = -1 × 7 = -7
For n=2
a2 = (-1)2 × 7(2)2 = 1 × 28 = 28
For n=3
a3 = (-1)3 × 7(3)2 = -1 × 63 = -63
Result
First three terms
-7, 28, -63
3.An expression for the nth triangular number is \frac{n(n+1)}{2}. Write down the 15th triangular number by taking n = 15.
Formula
Triangular number formula
an = n(n+1)2
Working
For n=15
a15 = 15(15+1)2 = 15 × 162
= 2402
= 120
Result
15th triangular number
120