The first term of an A.P is 3 and the fifth term is 9.Find the number of
terms in the progression if the sum of the progression is 81.
there are 4 differences between 3 and 9 ... d = (9 - 3) / 4 = 3/2
for n terms , the sum is ... 3 n + [d (n - 1) (n / 2)] = d/2 n^2 + (3 - d/2) n = 81
3/4 n^2 + 9/4 n - 81 = 0
use the quadratic formula to find n
n/2 (3+(n-1)(3/2)) = 81
1/4 (3n^2+3n-324) = 0
To find the number of terms in the arithmetic progression (A.P.), we need to use the formula for the sum of an arithmetic progression, which is given by:
S = (n/2)(2a + (n-1)d),
where S is the sum of the progression, a is the first term, n is the number of terms, and d is the common difference.
Given that the first term, a, is 3 and the sum of the progression, S, is 81, we can substitute these values into the formula to solve for n.
81 = (n/2)(2(3) + (n - 1)d).
Now, we need to find the common difference, d. We can do this by using the fifth term.
The fifth term of the arithmetic progression can be found using the formula:
a_n = a + (n - 1)d,
where a_n is the nth term.
Given that the fifth term, a_5, is 9 and the first term, a, is 3, we can substitute these values into the formula to solve for d.
9 = 3 + (5 - 1)d.
Now, we can solve this equation for d:
9 = 3 + 4d,
6 = 4d,
d = 1.5.
Now, we have the values of a (3), d (1.5), and S (81). Let's substitute them back into the formula for the sum of an arithmetic progression and solve for n:
81 = (n/2)(2(3) + (n - 1)(1.5)),
81 = (n/2)(6 + 1.5n - 1.5),
81 = (n/2)(4.5 + 1.5n),
81 = (n/2)(1.5n + 4.5).
Multiplying both sides of the equation by 2 to get rid of the fraction:
162 = n(1.5n + 4.5),
162 = 1.5n^2 + 4.5n,
1.5n^2 + 4.5n - 162 = 0.
Now, let's solve this quadratic equation. We can either factor it or use the quadratic formula:
Using the quadratic formula:
n = (-b ± √(b^2 - 4ac)) / (2a),
where a = 1.5, b = 4.5, and c = -162.
n = (-4.5 ± √((4.5)^2 - 4(1.5)(-162))) / (2(1.5)).
n = (-4.5 ± √(20.25 + 972)) / 3,
n = (-4.5 ± √(992.25)) / 3,
n = (-4.5 ± 31.49) / 3.
Simplifying the expression:
n = (26.99 / 3) or (-36.99 / 3).
Since the number of terms cannot be negative, we ignore the negative solution.
Therefore, the number of terms in the arithmetic progression is approximately:
n ≈ 26.99 / 3 ≈ 9.
So, there are approximately 9 terms in the progression.
To find the number of terms in the arithmetic progression (A.P.), we can use the formula for the sum of an A.P., which is given by:
\[ S_n = \frac{n}{2}(a_1 + a_n) \]
where S_n is the sum of the first n terms, a_1 is the first term, a_n is the nth term, and n is the number of terms.
Given that the first term (a_1) of the A.P. is 3, and the fifth term (a_5) is 9, we can substitute these values into the formula to get:
\[ S_n = \frac{n}{2}(3 + a_n) \]
We are also given that the sum of the A.P. (S_n) is 81. Substituting this into the equation, we have:
\[ 81 = \frac{n}{2}(3 + a_n) \]
Since we know the first term (a_1) is 3, and the fifth term (a_5) is 9, we can use the formula for the nth term of an A.P. to find the common difference (d):
\[ a_n = a_1 + (n-1)d \]
Substituting the values of a_1 and a_5, we get:
\[ 9 = 3 + 4d \]
Simplifying this equation gives:
\[ 4d = 6 \]
\[ d = \frac{6}{4} = \frac{3}{2} \]
Now, we can substitute this value of d back into the equation for S_n:
\[ 81 = \frac{n}{2}(3 + 3 + \frac{3}{2}(n-1)) \]
Simplifying, we have:
\[ 81 = \frac{n}{2}(6 + \frac{3n}{2} - \frac{3}{2}) \]
\[ 81 = \frac{n}{2}(\frac{12 + 3n - 3}{2}) \]
\[ 2(81) = n(6 + 3n - 3) \]
\[ 162 = n(3n + 3) \]
\[ 162 = 3n^2 + 3n \]
\[ 3n^2 + 3n - 162 = 0 \]
This is a quadratic equation. We can solve it by factoring or using the quadratic formula.
Factoring, we have:
\[ 3(n^2 + n - 54) = 0 \]
\[ n^2 + n - 54 = 0 \]
\[ (n + 9)(n - 6) = 0 \]
Setting each factor equal to zero, we get:
n + 9 = 0 or n - 6 = 0
n = -9 or n = 6
Since the number of terms (n) cannot be negative, we discard the solution n = -9. Therefore, the number of terms in the arithmetic progression is:
n = 6
So, the number of terms in the progression is 6.