the sum of the first 16th terms of an a.p is 240 and the sum of the next 4 terms is 220,find the first term
Let's break down the problem step-by-step to find the first term of the arithmetic progression (AP).
Step 1: Find the sum of the first 16 terms
The sum of an arithmetic progression can be calculated using the formula Sn = (n/2)(2a + (n-1)d), where Sn is the sum, n is the number of terms, a is the first term, and d is the common difference.
Given that the sum of the first 16 terms (Sn) is 240, and the number of terms (n) is 16, we have:
240 = (16/2)(2a + (16-1)d)
Simplifying this equation gives: 240 = 8(2a + 15d)
Or, 30 = 2a + 15d .......(Equation 1)
Step 2: Find the sum of the next 4 terms
Given that the sum of the next 4 terms is 220, and the number of terms (n) is 4, we have:
220 = (4/2)(2a + (4-1)d)
Simplifying this equation gives:
220 = 2(2a + 3d)
Or, 110 = 2a + 3d .......(Equation 2)
Step 3: Solving the equations
We have two equations:
Equation 1: 30 = 2a + 15d
Equation 2: 110 = 2a + 3d
We can solve these equations simultaneously to find the values of a and d.
Multiplying equation 2 by 5 gives: 550 = 10a + 15d
Subtracting equation 1 from this gives:
550 - 30 = (10a + 15d) - (2a + 15d)
520 = 8a
Dividing by 8 on both sides gives:
a = 520/8
a = 65
Thus, the first term (a) of the arithmetic progression is 65.
To find the first term (a) of an arithmetic progression (A.P.), you can use the formula for the sum of the terms of an A.P.
The sum of the first n terms of an arithmetic progression is given by the formula: Sn = (n/2)(2a + (n-1)d), where:
- Sn is the sum of the first n terms,
- a is the first term of the A.P.,
- n is the total number of terms, and
- d is the common difference between consecutive terms.
In this case, we are given the sum of the first 16 terms (Sāā) and the sum of the next 4 terms (Sāā).
Step 1: Find the sum of the first 16 terms.
Given: Sn = 240 and n = 16
240 = (16/2)(2a + (16-1)d)
240 = 8(2a + 15d)
Step 2: Find the sum of the next 4 terms.
Given: Sn = 220 and n = 4
220 = (4/2)(2a + (4-1)d)
220 = 2(2a + 3d)
Step 3: Solve the two equations simultaneously to find a and d.
8(2a + 15d) = 240 ... Equation 1
2(2a + 3d) = 220 ... Equation 2
Simplify Equation 1 by dividing both sides by 8:
2a + 15d = 30
Multiply Equation 2 by 4 to eliminate the coefficient 2 on the left side:
4a + 6d = 440
Now we have a system of linear equations:
2a + 15d = 30 ... Equation 3
4a + 6d = 440 ... Equation 4
To solve this system, we can use the method of substitution or elimination. I will use the elimination method.
Multiply Equation 3 by 2 to eliminate the coefficient 2 on the left side:
4a + 30d = 60 ... Equation 5
Now, subtract Equation 5 (4a + 30d = 60) from Equation 4 (4a + 6d = 440):
(4a - 4a) + (6d - 30d) = (440 - 60)
-24d = 380
d = 380 / -24
d = -15.83 (approximately)
Step 4: Substitute the value of d into Equation 3 or 4 to find the value of a.
Substituting d = -15.83 into Equation 3:
2a + 15(-15.83) = 30
2a - 237.45 = 30
2a = 30 + 237.45
2a = 267.45
a = 267.45 / 2
a = 133.73 (approximately)
Therefore, the first term of the A.P. is approximately 133.73.
16/2 (2a+15d) = 240
4/2 (2(a+16d) + 3d) = 220