1)The seventh term of an arithmetic progression is 15 while twice the third is 94.Calculate the first term and the common difference of the progression.
2)The first term of an arithmetic progression is 3 and the sum of its 8 terms is 164:
a)Find the common difference of the arithmetic progression.
b)Given that the sum of the first n terms of AP is 570,find n.
3)The first,the fifth and the seventh terms of another arithmetic sequence forms a decreasing geometric progression. If the first terms of the geometric progression is 64:
a)Find the values of the common differences of AP.
b)Find the first sum of the first ten terms of the GP.
4)In a geometric progression the sum of the second and third term is 6;and the sum of the third and fourth terms is 12.Find;
a)The first term
b)The common ratio.
c)The sum of number of the consecutive terms of an arithmetic progression is -19 and 1/2; the first term is 16 and 1/2; and the common difference is -3.Find the number of terms.
5)A factory manufacturers 100000 sweets in the first hour.The rate of production decreases by 20% every hour;
a)Calculate the number of sweets produced in the fourth hour.
b) Calculate the total number of sweets produced in the first 5 hours.
6)The eleventh term of an arithmetic progression is 4 times its second term.The sum of the first seven terms of the same progression is 175.Find the first term and the common difference of the progression.
7)Given the series 3+9+15+21+27+.......,find the number of the terms that will give a sum of 432.
1) Let the first term of the arithmetic progression be a and the common difference be d.
The seventh term is given by: a + 6d = 15
Twice the third term is given by: 2(a + 2d) = 94
From the first equation, we can solve for a in terms of d: a = 15 - 6d
Substituting this value into the second equation:
2(15 - 6d + 2d) = 94
30 - 12d + 4d = 94
-8d = 64
d = -8
Substituting this value of d back into the first equation:
a + 6(-8) = 15
a - 48 = 15
a = 63
The first term of the arithmetic progression is 63 and the common difference is -8.
2) Let the common difference of the arithmetic progression be d.
The sum of the arithmetic progression is given by: (8/2)(2a + 7d) = 164
Substituting the value of a = 3:
(4)(6 + 7d) = 164
24 + 28d = 164
28d = 140
d = 5
a) The common difference of the arithmetic progression is 5.
b) The sum of the first n terms of the arithmetic progression can be expressed as: (n/2)(2a + (n-1)d).
Substituting the given sum of 570, we have: (n/2)(2(3) + (n-1)(5)) = 570
(n/2)(6 + 5n - 5) = 570
(n/2)(5n + 1) = 570
5n^2 + n - 1140 = 0
(n - 12)(5n + 95) = 0
n = 12 or n = -95
Since the number of terms cannot be negative, n = 12.
The number of terms in the arithmetic progression is 12.
3) Let the first term of the arithmetic progression be a and the common difference be d.
The first term of the geometric progression is 64, so the common ratio of the geometric progression is r = (a + 4d)/a.
Since the fifth term of the arithmetic progression is the first term of the geometric progression, we have: a + 4d = 64
Since the seventh term of the arithmetic progression is the second term of the geometric progression, we have: a + 6d = 64r
Solving these two equations simultaneously:
Subtracting the second equation from the first equation, we get: 2d = 64 - 6d
8d = 64
d = 8
Substituting this value of d back into the first equation:
a + 4(8) = 64
a + 32 = 64
a = 32
The first term of the arithmetic progression is 32 and the common difference is 8.
To find the sum of the first ten terms of the geometric progression, we can use the formula for the sum of a geometric series:
S = a(1 - r^n)/(1 - r)
Substituting the values, S = 64(1 - (32/64)^10)/(1 - 32/64)
Simplifying, S = 64(1 - (1/2)^10)/(1/2)
S = 64(1 - 1/1024)/(1/2)
S = 64(1023/1024)/(1/2)
S = 64(1023/1024)(2)
S = 131328
The sum of the first ten terms of the geometric progression is 131328.
4) Let the first term of the geometric progression be a and the common ratio be r.
The second term is ar, the third term is ar^2, and the fourth term is ar^3.
From the given information, we have: ar + ar^2 = 6 and ar^2 + ar^3 = 12
Dividing the second equation by the first equation, we get: r^2 = 2
Taking the square root of both sides, we have: r = ±√2
Since the common ratio of a geometric progression cannot be negative, we take r = √2.
Substituting this value of r into the first equation:
a√2 + a(2) = 6
a(√2 + 2) = 6
a = 6/(√2 + 2)
Rationalizing the denominator, we get: a = 3(√2 - 2)
a) The first term of the geometric progression is 3(√2 - 2) and the common ratio is √2.
b) To find the sum of n terms of the geometric progression, we use the formula for the sum of a geometric series: Sn = a(1 - r^n)/(1 - r)
Substituting the given sum of -19 and the formula for the sum, we have:
-19 = (3(√2 - 2))(1 - (√2)^n)/(1 - √2)
Solving for n using algebraic manipulation, we find that n = 3.
The number of terms in the geometric progression is 3.
5) The rate of production decreases by 20% every hour, which means the number of sweets produced each hour is 80% of the previous hour.
a) The number of sweets produced in the fourth hour is: 100000(0.8)^3 ≈ 51200
b) The total number of sweets produced in the first 5 hours can be calculated by summing the geometric progression: 100000(1 - (0.8)^5)/(1 - 0.8) ≈ 100000(1 - 0.32768)/0.2 ≈ 335360
The total number of sweets produced in the first 5 hours is approximately 335360.
6) Let the first term of the arithmetic progression be a and the common difference be d.
The eleventh term is given by: a + 10d = 4(a + d)
Expanding and solving for a in terms of d: a = 3d
Substituting this value into the equation for the sum of the first seven terms:
(7/2)(2a + 6d) = 175
(7/2)(2(3d) + 6d) = 175
(7/2)(6d + 6d) = 175
21d = 175
d = 175/21 = 25/3
Substituting this value back into the equation for a = 3d:
a = 3(25/3) = 25
The first term of the arithmetic progression is 25 and the common difference is 25/3.
7) The series 3, 9, 15, 21, 27, ... is an arithmetic progression with a common difference of 6.
Let the number of terms be n.
The sum of the terms of an arithmetic progression can be calculated using the formula: S = (n/2)(2a + (n-1)d)
Substituting the given sum of 432 and the common difference of 6:
432 = (n/2)(2(3) + (n-1)(6))
432 = (n/2)(6 + 6n - 6)
432 = (n/2)(6n)
432 = 3n^2
n^2 = 432/3
n^2 = 144
n = ±12
Since the number of terms cannot be negative, n = 12.
The number of terms that will give a sum of 432 is 12.
1) Let's assume the first term of the arithmetic progression is "a" and the common difference is "d".
Given:
7th term (n = 7): 15
Twice the 3rd term (n = 3): 94
To find "a" and "d", we can use the formulas for the nth term of an arithmetic progression and twice the nth term:
nth term = a + (n-1)d
Using these formulas, we can set up the following equations:
15 = a + 6d ...(Equation 1)
94 = 2(a + 2d) ...(Equation 2)
Let's solve the equations simultaneously:
From Equation 1: a = 15 - 6d
Substituting this value in Equation 2:
94 = 2(15 - 6d + 2d)
94 = 30 - 8d
8d = 30 - 94
8d = -64
d = -8
Substituting d = -8 in Equation 1:
a = 15 - 6(-8) = 15 + 48 = 63
Therefore, the first term (a) is 63 and the common difference (d) is -8.
2) a) Given:
First term (a) = 3
Sum of 8 terms = 164
To find the common difference (d), we can use the formula for the sum of an arithmetic progression:
Sum of n terms = (n/2)(2a + (n-1)d)
Plugging in the known values:
164 = (8/2)(2(3) + (8-1)d)
164 = 4(6 + 7d)
41 = 6 + 7d
7d = 41 - 6
7d = 35
d = 5
Therefore, the common difference (d) is 5.
b) Given:
Sum of n terms (S) = 570
To find the value of n, we can use the same formula for the sum of an arithmetic progression:
570 = (n/2)(2a + (n-1)d)
Since we already know the first term (a) and common difference (d) from part a), we can substitute these values into the equation:
570 = (n/2)(2(3) + (n-1)(5))
570 = (n/2)(6 + 5n - 5)
570 = (n/2)(5n + 1)
Multiplying both sides by 2:
1140 = n(5n + 1)
5n^2 + n - 1140 = 0
This is a quadratic equation. Solving for n, we find that n ≈ 18.675 or n ≈ -3.675. Since we cannot have a negative number of terms, the number of terms (n) is approximately 19 for the sum to be 570.
3) Given:
First term (a) = 64
a) Let's assume the common difference of the arithmetic progression is "d".
The first, fifth, and seventh terms form a decreasing geometric progression. This implies:
a * (1/d) = a - 4d
a * (1/d^2) = a - 6d
To simplify this, we can multiply both equations by d^2:
a = ad^2 - 4d^3 ...(Equation 1)
a = ad - 6d^2 ...(Equation 2)
Substituting a = 64 in both equations:
64 = 64d^2 - 4d^3 ...(Equation 1')
64 = 64d - 6d^2 ...(Equation 2')
Let's solve these equations simultaneously:
From Equation 1': 4d^3 - 64d^2 + 64 = 0
d^3 - 16d^2 + 16 = 0 ...(Equation 3)
From Equation 2': 6d^2 - 64d + 64 = 0 ...(Equation 4)
Using synthetic division or polynomial division, we find that d = 1 is a solution to both equations 3 and 4.
Factoring the equations:
(d - 1)(d^2 - 15d + 16) = 0 ...(Equation 3')
(2d - 8)(3d - 8) = 0 ...(Equation 4')
This gives us two additional solutions: d = 8/2 = 4 and d = 8/3.
Therefore, the values of the common differences of the arithmetic progression are 1, 4, and 8/3.
b) To find the sum of the first ten terms of the geometric progression, we can use the formula:
Sum of n terms (S) = a * (1 - r^n)/(1 - r)
Given:
First term (a) = 64
To find the sum of the first ten terms, we need to know the common ratio (r) of the geometric progression.
Unfortunately, the problem does not provide information about the common ratio (r), so we cannot find the first sum of the first ten terms of the GP.