AN AP has 21 term the sum of 10 ,11,12 term is 129 and sum of last 3 term is 237.find AP
T10+T11+T12 = (a+9d)+(a+10d)+(a+11d), so
3a+30d = 129
Similarly, for the last three terms,
3a+57d = 237
a=3
d=4
and the sequence is
3,7,11,...,83
To find the arithmetic progression (AP), we need to determine the first term (a) and the common difference (d) of the sequence.
Let's start by finding the sum of the 10th, 11th, and 12th terms using the formula for the sum of an AP:
Sum = (n/2) * (2a + (n-1)d)
Where n is the number of terms, a is the first term, and d is the common difference.
Given that the sum of the 10th, 11th, and 12th terms is 129, we have:
129 = (3/2) * (2a + 2d)
Multiplying both sides by 2/3:
86 = 2a + 2d
1st equation: 2a + 2d = 86
Next, let's find the sum of the last three terms, which is given as 237:
237 = (3/2) * (2a + 40d) [since there are 21 terms and we need to consider the last three]
Multiplying both sides by 2/3:
158 = 2a + 40d
2nd equation: 2a + 40d = 158
Now we have a system of two simultaneous equations with two variables (a and d):
Equation 1: 2a + 2d = 86
Equation 2: 2a + 40d = 158
To solve this system of equations, we can subtract equation 1 from equation 2:
(2a + 40d) - (2a + 2d) = 158 - 86
This simplifies to:
38d = 72
Dividing both sides by 38:
d = 72/38
Simplifying further:
d = 36/19
Now, substitute the value of d back into equation 1 to find the value of a:
2a + 2(36/19) = 86
Multiply through by 19 to eliminate the fraction:
38a + 72 = 1634
38a = 1634 - 72
38a = 1562
Dividing both sides by 38:
a = 1562/38
a = 41
Therefore, the first term (a) of the AP is 41 and the common difference (d) is 36/19.
So, our arithmetic progression is: 41, (41 + 36/19), (41 + 2(36/19)), ..., (41 + 20(36/19)).
To find the arithmetic progression (AP), we need to determine the first term (a) and the common difference (d).
Let's consider the given information:
1. Sum of the 10th, 11th, and 12th term is 129:
We know that the sum of n terms in an AP can be calculated using the formula Sn = (n/2) * (2a + (n-1)d), where Sn is the sum of n terms, a is the first term, and d is the common difference.
Using the given information, we can write the equation:
(3/2)(2a + 2d) = 129 ....(1)
2. Sum of the last three terms is 237:
Using the same formula, we can write another equation for this information:
(3/2)(2a + 2d) + (3/2)(2a + d) + (3/2)(2a) = 237 ....(2)
Now, we can solve these two equations simultaneously to find the values of a and d.
Let's simplify equation (1) by multiplying both sides by 2/3:
(2a + 2d) = (2/3) * 129
2a + 2d = 86 ....(3)
Simplifying equation (2) further:
(2a + 2d) + (2a + d) + (3a) = 158 ....(4)
Now, let's solve equations (3) and (4) simultaneously:
Substituting the value of (2a + 2d) from equation (3) into equation (4):
86 + (2a + d) + (3a) = 237
86 + 5a + d = 237
Rearranging the terms:
5a + d = 151 ....(5)
Now, we have a system of equations (3) and (5):
2a + 2d = 86 ....(3)
5a + d = 151 ....(5)
Solving these equations simultaneously, we find:
a = 7
d = 36
Thus, the arithmetic progression (AP) is: 7, 43, 79, 115, 151, 187, 223, 259, 295, 331, 367, ...