The fourth and ninth terms of an arithmetic progression are -3 and 12 respectively. Find the:
1) common difference
2) fifth term
3) number of term which will give a sum of 135
1) To find the common difference, we can use the formula for the nth term of an arithmetic progression:
nth term = a + (n-1)d
where "a" is the first term, "n" is the number of the term we want to find, and "d" is the common difference.
We are given that the fourth term is -3 and the ninth term is 12. Plugging these values into the formula, we get:
-3 = a + (4-1)d
12 = a + (9-1)d
Simplifying these equations, we have:
-3 = a + 3d
12 = a + 8d
Next, we can subtract the first equation from the second equation:
12 - (-3) = (a + 8d) - (a + 3d)
Simplifying this, we get:
15 = 5d
Dividing both sides by 5, we find:
d = 3
Therefore, the common difference is 3.
2) To find the fifth term, we can use the formula we derived earlier:
nth term = a + (n-1)d
Plugging in the values we know, we have:
fifth term = a + (5-1)d
Substituting the value of the common difference, we get:
fifth term = a + (4)(3)
Simplifying this equation, we find:
fifth term = a + 12
Since we don't have the information to determine the value of "a," we can't find the exact value of the fifth term.
3) To find the number of the term that will give a sum of 135, we can use the formula for the sum of the first n terms of an arithmetic progression:
Sum = (n/2)(2a + (n-1)d)
Plugging in the values we know and the given sum, we have:
135 = (n/2)(2a + (n-1)d)
Again, since we don't have the information to determine the value of "a," we can't find the exact number of the term that will give a sum of 135.
To find the common difference of an arithmetic progression, we can subtract any term from the term preceding it. Let's subtract the fourth term from the ninth term:
12 - (-3) = 15
Therefore, the common difference is 15.
To find the fifth term, we can use the formula for the nth term of an arithmetic progression:
an = a1 + (n-1)d,
where:
an = the nth term,
a1 = the first term,
n = the position of the term,
d = the common difference.
Substituting the given values into the formula:
a5 = -3 + (5-1) * 15
a5 = -3 + 4 * 15
a5 = -3 + 60
a5 = 57
Therefore, the fifth term is 57.
To find the number of terms required to obtain a sum of 135 in this arithmetic progression, we can use the formula for the sum of an arithmetic series:
Sn = n/2 * (2a1 + (n-1)d),
where:
Sn = the sum of the first n terms,
a1 = the first term,
n = the number of terms,
d = the common difference.
Substituting the given values into the formula:
135 = n/2 * (2 * -3 + (n-1) * 15)
135 = n/2 * (-6 + 15n - 15)
135 = n/2 * (9n - 21)
Multiplying both sides by 2 to eliminate the fraction:
270 = n * (9n - 21)
270 = 9n^2 - 21n
Rearranging the equation:
9n^2 - 21n - 270 = 0
Factoring the quadratic equation, we get:
(3n + 9)(3n - 30) = 0
Setting each factor equal to zero:
3n + 9 = 0
n = -3/3
n = -1 (excluding this negative value, as we are dealing with the number of terms)
3n - 30 = 0
n = 30/3
n = 10
Therefore, the number of terms required to obtain a sum of 135 is 10.
To find the common difference of the arithmetic progression, we can use the formula:
nth term = a + (n-1)d
where 'a' is the first term, 'n' is the term number, and 'd' is the common difference.
1) Common Difference:
Given that the fourth term is -3, we can substitute the values into the formula:
-3 = a + (4-1)d
Similarly, given that the ninth term is 12, we can substitute the values into the formula:
12 = a + (9-1)d
We now have a system of two equations with two unknowns (a, d). We can solve this system of equations to find the values of 'a' and 'd'.
From the first equation, we can simplify:
-3 = a + 3d (equation 1)
From the second equation, we can simplify:
12 = a + 8d (equation 2)
Now, let's solve the system of equations using either substitution or elimination method.
Method 1: Substitution
From equation 1, we can isolate 'a':
a = -3 - 3d (equation 3)
Substituting equation 3 into equation 2, we have:
12 = (-3 - 3d) + 8d
12 = -3 + 5d
5d = 15
d = 3
So, the common difference (d) is 3.
2) Fifth Term:
To find the fifth term, we use the formula:
nth term = a + (n-1)d
Substituting the values, we have:
5th term = a + (5-1)d
Now, substitute the common difference (d) we found in step 1 and solve for 'a':
5th term = a + 4(3)
5th term = a + 12
Since we don't have the value of 'a', we cannot find the specific numerical value of the fifth term without more information.
3) Number of terms for a sum of 135:
To find the number of terms required to reach a sum of 135, we use the formula for the sum of an arithmetic series:
Sn = (n/2)(2a + (n-1)d)
Given the sum (Sn) is 135, we can iterate through different values of 'n' until we find the solution.
For each value of 'n', we can substitute the known values of 'a' and 'd' into the formula, then solve for 'n':
135 = (n/2)(2a + (n-1)d)
You can solve this equation by plugging in different values of 'n' until you get an equation that equals 135. Alternatively, you can use algebraic techniques such as factorization or solving the quadratic equation obtained from simplifying the equation above.