The fifth term of an AP is 82 and the 12th term is 103.find the first term and the common difference.
The sum of the first 21 terms
T12 = T5 + 7d, so d = (103-82)/7 = 3
Now you know a = T5-4d = 82-4*3 = 70
So, S21 = 21/2 (2*70 + 20*3) = 2100
S21=21/2(2(70)+(21-1)3
10.5(2(70)+(20)3)
10.5(140+60)
10.5(200)
=2100
To find the first term (a) and the common difference (d) of an arithmetic progression (AP), we can use the formulas:
1) nth term of an AP:
Tn = a + (n - 1)d
2) Sum of an AP:
Sn = (n/2)(2a + (n - 1)d)
Using the given information:
Given:
5th term (T5) = 82
12th term (T12) = 103
We can use the first formula to create two equations using the given information:
Equation 1:
82 = a + (5 - 1)d
Equation 2:
103 = a + (12 - 1)d
Simplifying Equation 1, we get:
82 = a + 4d ...(equation 3)
Simplifying Equation 2, we get:
103 = a + 11d ...(equation 4)
To solve these two equations, we can use the method of elimination or substitution. Let's use the method of substitution:
From equation 3, we have:
a = 82 - 4d
Now substitute the value of a in Equation 4:
103 = (82 - 4d) + 11d
Simplifying further:
103 = 82 + 7d
7d = 103 - 82
7d = 21
d = 21/7
d = 3
Now substitute the value of d in Equation 3 to find a:
82 = a + 4(3)
82 = a + 12
a = 82 - 12
a = 70
Therefore, the first term (a) is 70 and the common difference (d) is 3.
Now, to find the sum of the first 21 terms (S21), we can use the formula for the sum of an AP:
Sn = (n/2)(2a + (n - 1)d)
Substitute the given values:
n = 21
a = 70
d = 3
S21 = (21/2)(2(70) + (21 - 1)(3))
S21 = (21/2)(140 + 20(3))
S21 = (21/2)(140 + 60)
S21 = (21/2)(200)
S21 = (21)(100)
S21 = 2100
Therefore, the sum of the first 21 terms is 2100.
To find the first term (a) and the common difference (d) of an arithmetic progression (AP), use the given information:
1. The fifth term of the AP is 82:
a + 4d = 82 (since the fifth term is a + 4d)
2. The twelfth term of the AP is 103:
a + 11d = 103 (since the twelfth term is a + 11d)
Now, we have two equations with two variables (a and d). We can solve this system of equations to find the values of a and d.
Subtract equation 1 from equation 2 to eliminate 'a':
(a + 11d) - (a + 4d) = 103 - 82
11d - 4d = 21
7d = 21
d = 3
Now, substitute the value of d into equation 1 or 2 to find 'a':
a + 4(3) = 82
a + 12 = 82
a = 82 - 12
a = 70
Therefore, the first term (a) of the AP is 70 and the common difference (d) is 3.
To find the sum of the first 21 terms of the AP, we can use the formula for the sum of an AP:
Sum of n terms (S) = (n/2)(2a + (n-1)d)
Substituting the values we found:
S = (21/2)(2*70 + (21-1)*3)
S = (21/2)(140 + 20*3)
S = (21/2)(140 + 60)
S = (21/2)(200)
S = 2100
Therefore, the sum of the first 21 terms of the AP is 2100.