1.6.2 Sequence/Series Problems from Past Papers
1. The sum of the first n terms of an arithmetic sequence is Sn = 3n2 – 2n. Find the nth term un.
2. The nth term, un, of a geometric sequence is given by un = 3(4)n+1, n ∈Z+.
(a)
Find the common ratio r.
(b)
Hence, or otherwise, find Sn, the sum of the first n terms of this sequence.
3. Consider the arithmetic series 2 + 5 + 8 +....
(a)
Find an expression for Sn, the sum of the first n terms.
(b)
Find the value of n for which Sn = 1365.
4. A sequence {un} is defined by u0 = 1, u1 = 2, un+1 = 3un – 2un-1 where n ∈ Z+.
(a)
Find u2,u3,u4.
(b)
(i) Express un in terms of n.
(ii) Verify that your answer to part (b)(i) satisfies the equation
un+1 = 3un – 2un-1.
5. A geometric sequence has all positive terms. The sum of the first two terms is 15 and the
sum to infinity is 27. Find the value of
(a)
the common ratio;
(b)
the first term.
6. The first four terms of an arithmetic sequence are 2, a – b, 2a +b + 7, and a – 3b, where a
and b are constants. Find a and b.
7. The three terms a, 1, b are in arithmetic progression. The three terms 1, a, b are in
geometric progression. Find the value of a and of b given that a ≠ b.
8. Let {un}, n∈Z+, be an arithmetic sequence with the first term equal to a and common difference of
d, where d ≠ 0. Let another sequence {vn}, n∈Z+, be defined by
a.
i.
ii.
iii.
Show that
is a constant.
Write down the first term of the sequence {vn}.
Write down a formula for vn in terms of a, d, and n.
.
Let Sn be the sum of the first n terms of the sequence {vn}
b.
i.
Find Sn, in terms of a, d, and n
ii.
Find the values of d for which
exists.
You are now told that
does exist and is denoted by S∞.
iii.
Write down S∞ in terms of a and d.
iv.
Given that S∞= 2a+1 find the value of d.
Let {wn}, n∈Z+, be a geometric sequence with first term equal to p and common ratio q, where p and q
are both greater than zero. Let another sequence {zn} be defined by zn=ln wn.
c. Find
, giving your answer in the form ln k with k in terms of n, p, and q.
9. Evaluate
a.
b.
Caution: telescoping series
c.
d.
Answers
1. un=6n-52a. r=4 b. Sn=16(4n-1) 3a. Sn=(n/2)(4+3(n-1)) b. n=30
4. a. 4,8,16
bi. un=2n ii. LS=2n+1 RS=3(2n)-2(2n-1), prove LS=RS
5. a. ⅔ b. 9
6. a=2, b=-3
7. a=-2, b=4
8. ai. vn=2d aii. v1=2a iii. vn=2a(2d)n-1 bi.
iii.
bii. d<0
iv. d=-1
c.
9a. 15550
b. 6/31
c. 11900
d. 1668550
0
