1.Find an A.P. such that the sum of the first three terms is one half the sum of the next four terms,the first term being 12.
2.In an A.P.,the sum of the first three terms is 18,and the sum of the squares of these terms is 126.Find the terms.
Just use the definitions of an AP and translate the English to Math.
"the sum of the first three terms"
= a + a+d + a+2d
= 3a + 3d
"the sum of the next four terms"
= a+3d + a+4d + a+5d
= 3a + 12d
given:
"the sum of the first three terms is one half the sum of the next four terms"
3a+3d = (1/2)(3a+12d)
times 2
6a+6d = 3a+12d
3a=6d
a = 2d
but a = 12, so d =6
the AP is 12, 18, 24, 30, 36, 42, ...
check:
sum of first 3 terms = 12+18+24 = 54
sum of next 3 terms = 30+36+42 = 108
is the first sum one half of the second sum ?
Try the second question in the same way, show me your steps
765
To solve these problems, we need to recall the formulas for the sum of an arithmetic progression (AP) and the sum of the squares of an AP.
1. Let's find the terms of the AP, where the sum of the first three terms is one half the sum of the next four terms, with the first term being 12.
First, let's denote the first term of the AP as "a" and the common difference as "d". So, the first term of the AP is 12 (given).
The sum of the first three terms can be calculated using the formula: S3 = (3/2) * a + (3/2) * d.
The sum of the next four terms can be calculated using the formula: S4 = (4/2) * a + (4/2) * d.
According to the given condition, S3 = (1/2) * S4.
Substituting the values, we get: (3/2) * a + (3/2) * d = (1/2) * [(4/2) * a + (4/2) * d].
Simplifying the equation, we have: 3a + 3d = a + d.
Simplifying further, we get: 2a + 2d = 0.
Dividing by 2, we have: a + d = 0.
Since the first term is 12 (given), we can substitute a = 12 into the equation and solve for the common difference:
12 + d = 0
d = -12
Now we have the first term (a = 12) and the common difference (d = -12). The terms of the AP are: 12, 0, -12, -24, -36, ...
2. Let's find the terms of the AP, where the sum of the first three terms is 18, and the sum of the squares of these terms is 126.
Using the standard formulas, the sum of the first three terms (S3) is given as: S3 = (3/2) * a + (3/2) * d = 18.
The sum of the squares of the first three terms (S3²) is given as: S3² = (a² + (a + d)² + (a + 2d)²) = 126.
Substituting the value of S3 in the equation, we get: (3/2) * a + (3/2) * d = 18.
Simplifying, we have: 3a + 3d = 36.
Substituting the value of S3² in the equation, we get: (a² + (a + d)² + (a + 2d)²) = 126.
Simplifying, we have: a² + (a² + 2ad + d²) + (a² + 4ad + 4d²) = 126.
After combining like terms, we have: 3a² + 6ad + 6d² = 126.
Now, we have a system of two equations with two variables:
3a + 3d = 36 -- (Equation 1)
3a² + 6ad + 6d² = 126 -- (Equation 2)
We can solve this system of equations to find the values of "a" and "d".