Download Solved Problems – Arithmetic and Geometric

Arithmetic/Geometric Sequence and Series
1. Find the values of x and y so that x, y and 5x are in arithmetic progression with the sum equal to
81.
Solution:
Use the definition of arithmetic progression.
y – x = 5x – y y = 3x (Eq. 1)
x + y + 5x = 81 (Eq. 2)
Solving for x and y, x = 9 and y = 27.
2. In a geometric progression, the first and the second terms are x-4 and xt respectively. If a8 = x52,
what is the value of t?
Solution:
r = xt/x-4
a8 = a1rn – 1 x52 = x-4(xt/x-4)7 x52 = x24 + 7t t = 4
3. A pendulum swings through an arc of 36 inches. On each successive swing, the pendulum covers
an arc equal to 90% of the previous swing. Find the length of the arc on the 7th swing and the
total distance the pendulum travels before coming to rest.
Solution:
(1)
(2)
a7 = a1r6 a7 = 36(0.9)6 = 19.13 inches
S = a1 / (1 – r) 36 / (1 – 0.9) = 360 inches
4. In an arithmetic sequence of positive numbers, the common difference is twice the first term
and the sum of the first six terms is equal to the square of the first term. Find the first term of
the sequence.
Solution:
Let a be the first term, then 2a is the common difference.
Use the formula S =
a2 =
n
[2a1 + (n − 1)d] .
2
6
[2a + (6 − 1)2a]
2
a2 = 36
a=6
5. A certain ball when dropped from a height rebounds 3/5 of the original height. How high will the
ball rebound at the 4th bounce if it is dropped from a height of 10 ft?
a4 = a1rn – 1 S = 10(3/5)3= 10(27/125) = 10 4/5 ft
or
54/5 ft
1
6. Two positive numbers may be inserted between 3 and 9 such that the first three numbers in the
sequence from a geometric progression and the last three form an arithmetic progression. Find
the sum of these two positive numbers?
Solution:
The sequence is 3, a, b, 9 where a, b > 0.
The first three numbers form a GP. a/3 = b/a a2 = 3b b = a2/3 (eq. 1)
The last three numbers form a AP. b – a = 9 – b 2b – a – 9 = 0 (eq. 2)
Solve equations (1) and (2).
2a2 – 3a – 27 = 0 (2a – 9)(a + 3) = 0 a = 9/2 or a = - 3. Reject a = -3, since a > 0.
Hence, b = 27/4. Thus, a + b = 9/2 + 27/4 = 45/4
7. The digits of a certain 3 – digit number form an arithmetic sequence. If their sum is 15 and the
number is 119 times the hundreds digit, what is the number?
Solution:
Let x – r be the hundreds digit
x be the tens digit
x + r be the units digit
(x – r) + x + (x + r) = 15 x =5
Number: 100(x – r) + 10x + (x + r) = 119(x – r), r = 2
Thus, the number is 357.
8. Find the values of m and n if (3m – n), (2m + n), (5m + 1), (6m + n) are consecutive terms of an
arithmetic progression.
Solution:
2m + n – (3m – n)= 5m + 1 – (2m + n) Eq. 1
5m + 1 – (2m + n) = 6m + n – (5m + 1) Eq. 2
Solving for m and n, m = 2 and n = 3.
9. The sum of the first six terms of a geometric progression is nine times the sum of its first three
terms. Find the common ratio.
Solution:
S6 = 9S3 a1 (1 − r 6 ) 9a1 (1 − r 3 )
=
1 – r6 = 9(1 – r3) (1 + r3)(1 – r3) = 9(1 – r3)
1 −r
1 −r
1 + r3 = 9 r = 2
2
10. In a sequence, a1 = 2, a2 = 3 and an = (an – 2)2 + an – 1 , find a5.
a3 = a12 + a2 = 22 + 3 = 7
a4 = a22 + a3 = 32 + 7 = 16
a5 = a32 + a4 = 72 + 16 = 65
11. Find an arithmetic progression whose first term is unity such that the second, tenth and thirty –
fourth terms form a geometric progression.
a1 = 1
a2 = 1 + d
a10 = 1 + 9d
Note that a2, a10 and a34 form a GP. Hence,
a34 = 1 + 33d
a10 a 34
.
=
a 2 a10
a10 a 34
1 + 9d 1 + 33d
=
(1 + 9d) 2 = (1 + d)(1 + 33d)
=
a 2 a10
1+d
1 + 9d
Simplifying, 48d2 – 16d = 0, d = 1/3
Therefore, a2 = 4/3 a10 = 4 and a34 = 12.
12. The common ratio of a geometric progression is 2/3. What percent of the sum to the infinity is
the sum of the first 4 terms of the geometric progression?
Solution:
a1 = a
a2 = ar
a3 = ar2
a4 = ar3
where r = 2/3
2
4
8
65
a+ a+ a = a
3
9
27
27
a
= 3a
Sinf =
1 − 23
S4 = a +
%=
S 4 65a 27
=
= 0.802 , Hence, the sum of the first 4 terms form 80% of the sum of infinity.
S inf
3a
13. The 1st and the 3rd terms of a geometric progression are x-4 and xm. If the 8th term is x52, what is
m? [ 12 ]
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