In an arithmetic progression, the 5th term is six times the 1st term and the sum of the first six terms is 99. Find the 1st term and the common difference.
Attempt at solution:
U5 = 6(U1)
99 = 3(2[U1] + 2d)
Just use your definitions.
Isn't the definition of 5th term (a + 4d)
and that of the first term a ?
so a+4d = 6a
4d = 5a
d = 5a/4 -----> **
Also : sum(6) = (6/2)(2a + 5d)
99 = 3(2a + 5(5a/4) )
33 = 2a + 25a/4
times 4
132 = 8a + 25a
continue to solve for a, then sub it back into **
Thanks so much I appreciate I now understand
Thanks for all your answers, I get it now
5. In an geometric progression the term is 432 and the 4th term is 128. Find the common ratio and the sum to infinity of the progression.
5. In an geometric progression the term is 432 and the 4th term is 128. Find the common ratio and the sum to infinity of the progression.
Well, aren't you finding all the terms and sums a bit *progressive*? I guess I'll have to put on my comedy hat and help you out. Let's solve this jolly puzzle!
We know that in an arithmetic progression, the nth term can be calculated using the formula: Un = U1 + (n - 1)d, where U1 represents the first term and d is the common difference.
So let's use this formula to solve our quandary. According to the information given, we have:
U5 = 6U1 ----(1)
U1 + U2 + U3 + U4 + U5 + U6 = 99 ----(2)
Let's substitute the formula for Un into equation (2):
U1 + (U1 + d) + (U1 + 2d) + (U1 + 3d) + (U1 + 4d) + (U1 + 5d) = 99
Simplifying this expression, we get:
6U1 + 15d = 99
From equation (1), we know that U5 = 6U1, so we can rewrite the equation as:
U5 + 15d = 99
Since U5 = U1 + 4d, we can now rewrite the equation one more time:
U1 + 4d + 15d = 99
Combining like terms, we get:
U1 + 19d = 99
Now, we have a system of two equations:
6U1 = U5 ----(1)
U1 + 19d = 99 ----(3)
Now we can solve this system of equations to find the values of U1 and d.
But before we dive into the calculations, let me tell you a joke to lighten the mood: Why don't scientists trust atoms? Because they make up everything!
Alright, let's proceed. Simplifying equation (1), we get:
6U1 - U1 = 5U1 = U5
Substituting this result into equation (3), we get:
5U1 + 19d = 99
Now, we have a linear equation with two variables. Let's keep the calculations going and solve for U1:
5U1 + 19d = 99
5U1 = 99 - 19d
U1 = (99 - 19d)/5
And there you have it! The formula for the first term U1. Now, you can plug in any value for d to find the corresponding U1. For example, if d = 2, then U1 = (99 - 19(2))/5 = 61/5.
But, before we conclude, what did one math book say to the other math book? I've got problems!
I hope my jests and explanations brought a smile to your face while solving this arithmetic progression riddle. If you have any more questions or need further clarification, feel free to ask, my humorous friend!
To solve this problem, we can use the formulas for the nth term and the sum of an arithmetic progression.
The nth term of an arithmetic progression can be determined using the formula:
Un = U1 + (n - 1)d
where Un is the nth term, U1 is the first term, n is the term number, and d is the common difference.
The sum of the first n terms of an arithmetic progression can be determined using the formula:
Sn = n/2 * (2U1 + (n - 1)d)
where Sn is the sum of the first n terms of the arithmetic progression.
Given that the 5th term (U5) is six times the first term (U1) and the sum of the first six terms (S6) is 99, we can form two equations.
Equation 1: U5 = 6U1
Equation 2: S6 = 99
We can substitute the formulas of Un and Sn into these equations to solve for U1 and d.
Let's solve the equations step by step:
From Equation 1, we have U5 = 6U1.
Substituting the formula of the nth term (Un = U1 + (n - 1)d), we get:
U1 + 4d = 6U1.
Rearranging, we have:
5U1 = 4d.
This is our first equation.
Now, let's solve Equation 2: S6 = 99.
Substituting the formula of the sum of the first n terms (Sn = n/2 * (2U1 + (n - 1)d)), we get:
6/2 * (2U1 + 5d) = 99.
Simplifying, we have:
3 * (2U1 + 5d) = 99,
6U1 + 15d = 99.
Now, let's substitute the value of 5U1 from our first equation into the second equation:
5U1 + 15d = 99.
Substituting 4d for 5U1, we get:
4d + 15d = 99,
19d = 99.
Dividing both sides by 19, we have:
d = 99/19 ≈ 5.21.
Now, let's substitute the value of d back into our first equation:
5U1 = 4d,
5U1 = 4 * 5.21,
5U1 = 20.84.
Dividing both sides by 5, we have:
U1 = 20.84/5 ≈ 4.17.
Therefore, the first term (U1) is approximately 4.17 and the common difference (d) is approximately 5.21.