Business Mathematics
Arithmetic Progression (AP) and Geometric
Progression (GP)
CONTENTS:
1. Formulas
2. 24 Selected Problems
3. Solutions to the Selected Problems
Arithmetic Progression Formulas
|
1 |
nth
term of AP, tn = |
a
+ (n – 1)d Where,
a = 1st term of the series, and d = common difference |
|
2 |
Arithmetic
Mean (AM) of a and c, M = |
½
(a + c) |
|
3 |
Sum
of n terms in AP, Sn = |
(i)
n/2[2a
+ (n – 1)d] |
|
4 |
Sum
of n terms in AP, Sn = |
(ii)
n/2(a
+ l) [l =
last term] |
|
5 |
Sum
of the first n natural numbers = |
[n(n
+ 1)]/2 |
|
6 |
Sum
of the first n odd natural numbers = |
n^2 |
|
7 |
Sum
of the first n even natural numbers = |
n(n
+ 1) |
|
8 |
Sum
of the squares of the first n natural numbers = |
[n(n
+ 1)(2n + 1)]/6 |
|
9 |
Sum
of the cubes of the first n natural numbers = |
[{n(n
+ 1)}/2]^2 |
Geometric Progression Formulas
|
1 |
nth
term of GP, tn = |
a[r^(n
– 1)] Where,
a = 1st term of the series, and r = common ratio |
|
2 |
Geometric
Mean (GM) of a and c, G = |
± (ac)^(½) |
|
3 |
Sum
of n terms in GP, Sn [Where, r <
1] = |
(i)
[a(1
– r^n)] / (1 –
r) |
|
4 |
Sum
of n terms in GP, Sn [Where, r <
1] = |
(ii)
(a
– rl) / (1 – r) [l
= last term] |
|
5 |
Sum
of n terms in GP, Sn [Where, r > 1]
= |
(i)
[a(r^n
– 1)] / (r – 1) |
|
6 |
Sum
of n terms in GP, Sn [Where, r > 1]
= |
(ii)
(rl
– a) / (r – 1) [l =
last term] |
Arithmetic
Progression and Geometric Progression
Selected
Problems
Arithmetic
Progression
1. Find the 7th
term of the A.P. 8, 5, 2, -1, -4.....
2. If 5th
and 12th terms of an A.P. are 14 and 35 respectively find the A.P.
3. Which term of
the A.P. 3/√7, 4/√7, 5/√7......is 17/√7.
4. Divide 69 into
three parts which are in A.P. and are such that the product of the 1st
two parts is 483.
5. Find the
arithmetic mean between 4 and 10.
6. Insert 4 arithmetic means between 4 and 324.
7. A man secures an interest free loan of Rs 14,500
from a friend and agrees to repay it in 10 instalments. He pays Rs 1,000 as
first instalment and then increases each instalment by equal amount over the
preceding instalment. What will be his last instalment?
8. Mr. X arranges to pay off a debt of Rs 9,600 in 48
annual instalments which form an AP. When 40 of these instalments are paid, Mr.
X becomes insolvent with Rs 2,400 still remaining unpaid. What is the value of
second instalment?
9. A man saved Rs 16,500 in 10 years. In each year
after the first he saved Rs 100 more than he did in the preceding year. What
was his savings in the first year?
10. Find the sum of all natural numbers from 100 to
300 which are divisible by 4.
11. The rate of monthly salary of a person is
increased annually in AP. It is known that he was drawing Rs 400 a month during
the 11th year of his service and Rs 760 during the 29th
year. Find his starting salary and the rate of annual increment. What should be
his salary at the time of retirement just on the completion of 36th
year of service?
12. A money lender lends Rs 1,000 and charges an
overall interest of Rs 140. He recovers the loan and interest by 12 monthly
instalments each of which after the first instalment being less by Rs 10 than
the preceding one. What was the amount of the first instalment?
Geometric Progression
1. If a, ar, ar^2,
ar^3........be in G.P. Find the common ratio.
2. Which term of
the progression 1, 2, 4, 8.......is 256?
3. Insert 3
geometric means between 1/9 and 9.
4. Find the G.P. where
4th term is 8 and 8th term is 128/625.
5. Find the sum of 1+2+4+8........up
to 8 terms.
6. Find three numbers in GP whose sum is 19 & product is 216.
7. Find the sum of the series (1/6 +
1/6^2 + 1/6^3...................1/6^n) [n
∞]
9. Three numbers are in A.P and their sum is 15. If 1, 3, 9 are added to them respectively, they form a G.P. Find the numbers.
10. Find the sum of the series 6, 27, 128, 629....up to 10th term.
11. Find the sum of
n terms of the series 3 + 33 + 333................
12. Find the sum of
n terms of the series 0.7 + 0.77 + 0.777............
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