Problem 6: 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.
Why did the arithmetic sequence go to therapy? Because it couldn't solve its problems!
To solve this problem, we can first find the common difference (d) of the arithmetic sequence using the given information.
We know that the sum of an arithmetic sequence can be calculated using the formula:
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.
Given that a1 + a3 + a5 + a7 + a9 = 17 and a2 + a4 + a6 + a8 + a10 = 15, we can substitute these values into the sum formula:
17 = (5/2)(2a + 4d) ...(1)
15 = (5/2)(2(a + d)) ...(2)
Now, let's simplify these equations:
Multiplying equation (1) by 2, we get:
34 = 10a + 20d
Multiplying equation (2) by 2, we get:
30 = 10a + 10d
Now, subtracting equation (2) from equation (1), we eliminate variable 'a' and solve for 'd':
34 - 30 = 10a + 20d - (10a + 10d)
4 = 20d - 10d
4 = 10d
d = 4/10
d = 0.4
Now that we have found the common difference (d = 0.4), we can find the value of a1 by substituting this value into equation (1) and solving for a:
17 = (5/2)(2a + 4d)
17 = (5/2)(2a + 4(0.4))
17 = (5/2)(2a + 1.6)
Expanding the expression:
17 = (5/2)(2a) + (5/2)(1.6)
17 = 5a + 4
Subtracting 4 from both sides:
17 - 4 = 5a + 4 - 4
13 = 5a
Dividing both sides by 5:
a = 13/5
a = 2.6
Therefore, the value of a1 is 2.6.
To solve this problem, we can use the concept of arithmetic sequences and their properties.
An arithmetic sequence is a sequence of numbers in which the difference between consecutive terms is constant. In this case, let's assume that the common difference between the terms is 'd'.
Now, let's break down the given information and simplify the expressions:
a1 + a3 + a5 + a7 + a9 = 17 (Equation 1)
a2 + a4 + a6 + a8 + a10 = 15 (Equation 2)
We can rewrite Equation 1 using the formula for the nth term of an arithmetic sequence:
a1 + (a1 + 2d) + (a1 + 4d) + (a1 + 6d) + (a1 + 8d) = 17
Simplifying this equation, we get:
5a1 + 20d = 17 (Equation 3)
Similarly, we rewrite Equation 2:
(a1 + d) + (a1 + 3d) + (a1 + 5d) + (a1 + 7d) + (a1 + 9d) = 15
Simplifying this equation, we get:
5a1 + 25d = 15 (Equation 4)
To find the value of 'a1', we can solve this system of equations (Equation 3 and Equation 4) using substitution or elimination.
Subtracting Equation 4 from Equation 3, we get:
20d - 25d = 17 - 15
-5d = 2
d = -2/5
Now, substitute the value of 'd' into Equation 3 or Equation 4 to solve for 'a1'. Let's use Equation 3:
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.