Let a1,a2,......a10 be an arithmetic sequence. If a1+a3+a5+a7+a9=17 and a2+a4+a6+a8+a10=15 then find a1.
A6, A7,A8,A9 and A10. A1 is 5, d is 2
Let's solve this step by step.
Step 1: Find the common difference (d) of the arithmetic sequence.
Since the sequence is arithmetic, the common difference will be the same between any two consecutive terms.
Step 2: Express the given sum using the formula for the sum of an arithmetic series.
The sum of an arithmetic series can be expressed as:
Sn = (n/2)(2a + (n-1)d)
where Sn is the sum of the first n terms, a is the first term, and d is the common difference.
In this case, the given sums can be expressed as:
17 = (5/2)(2a1 + (5-1)d) ...(1)
15 = (5/2)(2a1 + (5-1)d) ...(2)
Step 3: Simplify the equations.
Equations (1) and (2) can be simplified to:
34 = 10a1 + 10d ...(3)
30 = 10a1 + 10d ...(4)
Step 4: Solve the system of equations.
Subtracting equation (4) from equation (3), we get:
34 - 30 = 10a1 + 10d - (10a1 + 10d)
4 = 0
This equation is inconsistent, meaning there is no possible solution.
Therefore, there is no value for a1 that satisfies the given conditions.
To find the value of a1, we need to understand the properties of an arithmetic sequence and solve the given equations.
An arithmetic sequence is a sequence in which the difference between any two consecutive terms is constant. Let's denote the common difference as 'd'. So, for an arithmetic sequence, we have:
a2 = a1 + d
a3 = a1 + 2d
a4 = a1 + 3d
a5 = a1 + 4d
a6 = a1 + 5d
a7 = a1 + 6d
a8 = a1 + 7d
a9 = a1 + 8d
a10 = a1 + 9d
Using this information, we can rewrite the given equations as follows:
a1 + (a1 + 2d) + (a1 + 4d) + (a1 + 6d) + (a1 + 8d) = 17 ---> Equation 1
(a1 + d) + (a1 + 3d) + (a1 + 5d) + (a1 + 7d) + (a1 + 9d) = 15 ---> Equation 2
Now, let's substitute the expressions for a2, a3, a4, etc., into Equations 1 and 2:
5a1 + 20d = 17 ---> Equation 3
5a1 + 25d = 15 ---> Equation 4
To solve these equations, we'll subtract Equation 4 from Equation 3:
20d - 25d = 17 - 15
-5d = 2
d = -2/5
Now we know the value of 'd'. We can substitute it into Equation 3 to find the value of a1:
5a1 + 20(-2/5) = 17
5a1 - 8 = 17
5a1 = 17 + 8
5a1 = 25
a1 = 25/5
a1 = 5
Therefore, the value of a1 is 5.
a1+...+a9 = 5a + (0+2+4+6+8)d
a2+...+a10 = 5a + (1+3+5+7+9)d
so,
5a+20d = 17
5a+25d = 15
d = -2/5
a = 5