The sum of the first 10 terms of an Ap is 145 and the sum of its fourth and ninth is 5 times the third term determine the first term and the constant difference
S10=145
T4+T9=5×T3
therefore we can write this as
(a+3d)+(a+8d)=5(a+2d)
2a+11d=5a+10d
d=3a
to find the first term we can substitute our S10 & common difference into Sn=n/2[2a+(n-1)d] .
S10=10/2[2a+(10-1)3a]
145=5(2a+30a-3a)
145=5(29a)
145=145a {divide both sides by 145}
.•. a = 1
substitute a in d=3a to find the difference.
d=3(1)
d=3
10/2 (2a+9d) = 145
a+3d + a+8d = 5(a+2d)
Well, solving math problems can be a bit tricky, but I'll give it a shot! Here's what I think:
Let's assume that the first term of the arithmetic progression (AP) is 'a' and the constant difference is 'd'.
Now, the sum of the first 10 terms of an AP can be calculated using the formula:
Sum = (n/2) * (2a + (n-1)d)
Given that the sum is 145, we can plug in the values:
145 = (10/2) * (2a + (10-1)d)
145 = 5 * (2a + 9d)
29 = 2a + 9d --- Equation 1
We also know that the sum of the fourth and ninth terms is 5 times the third term. Let's denote the third term as 't'.
So, the fourth term would be 't + d', and the ninth term would be 't + 5d'.
According to the given condition, we have:
(t + d) + (t + 5d) = 5t
2t + 6d = 5t
6d = 3t
2d = t --- Equation 2
Now, we have two equations, Equation 1 and Equation 2, and we can solve them simultaneously to determine the values of 'a' and 'd'.
Substituting Equation 2 into Equation 1, we get:
29 = 2a + 9(2d)
29 = 2a + 18d
Dividing both sides by 2:
14.5 = a + 9d --- Equation 3
Now, we can solve Equations 2 and 3 simultaneously:
2d = 14.5 - 9d
11d = 14.5
d = 14.5/11
d ≈ 1.318181818...
Substituting the value of 'd' back into Equation 2:
t = 2d
t ≈ 2 * 1.318181818...
t ≈ 2.636363636...
Therefore, the first term (a) would be:
a = t - 2d
a ≈ 2.636363636... - 2 * 1.318181818...
a ≈ 0
So, by solving the equations, we find that the first term is approximately 0 and the constant difference is approximately 1.318181818... (which is a funny number to work with!).
I hope this helps, and remember, math jokes are always the best way to lighten the mood when dealing with equations!
To determine the first term and the constant difference of an arithmetic progression (AP) based on the given information, we can use the formulas for the sum of the first n terms of an AP and the nth term of an AP.
Let's denote the first term as 'a' and the constant difference as 'd'.
Given: The sum of the first 10 terms of the AP is 145.
Using the formula for the sum of the first n terms of an AP:
S_n = (n/2)(2a + (n-1)d)
Substituting the values, we have:
145 = (10/2)(2a + (10-1)d)
145 = 5(2a + 9d)
29 = 2a + 9d -- Equation 1
Given: The sum of the fourth and ninth terms is 5 times the third term.
Using the formula for the nth term of an AP:
a_n = a + (n-1)d
The fourth term is a + 3d, the ninth term is a + 8d, and the third term is a + 2d.
According to the given information:
(a + 3d) + (a + 8d) = 5(a + 2d)
2a + 11d = 5a + 10d
11d - 10d = 5a - 2a
d = 3a -- Equation 2
Now, we need to solve these two simultaneous equations (Equation 1 and Equation 2) to find the values of 'a' and 'd'.
Substituting the value of 'd' from Equation 2 into Equation 1:
29 = 2a + 9(3a)
29 = 2a + 27a
29 = 29a
a = 1
Substituting the value of 'a' back into Equation 2 to find 'd':
d = 3a
d = 3(1)
d = 3
Therefore, the first term is 1 and the constant difference is 3 in the given arithmetic progression (AP).