The sum of n terms of the sequence 3,9,15,21 is 7500 determine value if n
To find the value of n, we need to determine the pattern of the sequence and then find the general formula for the nth term.
From 3 to 9, we add 6.
From 9 to 15, we add 6.
From 15 to 21, we add 6.
Therefore, the sequence has a common difference of 6.
The formula for the nth term of an arithmetic sequence is given by: tn = a + (n - 1)d, where a is the first term and d is the common difference.
In this case, the formula is: tn = 3 + (n - 1)6.
Now, we need to find the value of n when the sum of the sequence is 7500.
The sum of an arithmetic series is given by the formula: Sn = (n/2)(a + tn), where Sn is the sum of the first n terms.
In this case, Sn = 7500 and a = 3. Substituting the values into the formula:
7500 = (n/2)(3 + (3 + (n - 1)6))
7500 = (n/2)(3 + 3 + 6n - 6)
7500 = (n/2)(6n)
7500 = 3n^2
2500 = n^2
n = √2500
n = 50 or n = -50
Since n represents the number of terms in the sequence, it cannot be negative. Therefore, the value of n is 50.
To determine the value of n, we need to find out how many terms are in the sequence 3, 9, 15, 21 until the sum reaches 7500.
Let's set up the equation to find the sum of n terms of an arithmetic sequence:
Sn = (n/2)(2a + (n-1)d)
Where:
Sn = sum of n terms
n = number of terms
a = first term
d = common difference
In this case, the first term is 3, and since the sequence is increasing by 6 each time, the common difference is 6.
So, Sn = (n/2)(2(3) + (n-1)(6))
= (n/2)(6 + 6n - 6)
= (n/2)(6n)
= 3n^2
Now we can solve for n by equating Sn to 7500:
3n^2 = 7500
n^2 = 2500
n = sqrt(2500)
n = 50
Therefore, the value of n is 50.