The 10th term of an arithmetic series is 34, and the sum of the first 20 terms is 710. Determine the 25th term.
Can someone please explain how to do this? I don't know which formula to use. TIA
You know t(10) = a+9d
a+9d = 34
or a = 34-9d
S(20) = 20/2[2a + 19d]
710 = 10(2a + 19d)
71 = 2a + 19d
71 = 2(34-9d) + 19d
71 = 68 - 18d + 19d
3 = d
a = 34-27 = 7
then t(25) = a+24d = 7 + 24(3) = 79
for t(10)= a+9d .. where did you get the 9 from? & can you explain how you got 710? thank you
oh sorry sorry. i never got that you simplified it. thanks
you should know that
t(n) = a + d(n-1) one of the most basic formulas in the study of sequences
and you gave me 710 in the question!
hi, i was wondering where you got:
20/2[2a + 19d] from, im in yr 11 but want to start revising a level because i want to get a phd in maths, so i just wanted to know where you got it from, if you could answer this, then thanks
This was GREAT
To find the 25th term of an arithmetic series, we need to use two pieces of information given in the problem: the 10th term and the sum of the first 20 terms.
Let's start by finding the common difference (d) of the arithmetic series. The formula for the nth term (Tn) of an arithmetic series is: Tn = a + (n - 1) * d, where a is the first term and d is the common difference.
We are given that the 10th term (T10) is 34. Plugging this value into the formula, we get: 34 = a + (10 - 1) * d. Simplifying this equation, we have: 34 = a + 9d.
Next, we need to find the sum of the first 20 terms (S20). The formula for the sum of an arithmetic series (S) is: S = (n/2) * (2a + (n - 1) * d), where n is the number of terms.
We are given that the sum of the first 20 terms (S20) is 710. Plugging this value into the formula, we get: 710 = (20/2) * (2a + (20 - 1) * d). Simplifying this equation, we have: 710 = 10 * (2a + 19d).
Now, we have a system of two equations with two unknowns (a and d):
Equation 1: 34 = a + 9d
Equation 2: 710 = 10 * (2a + 19d)
We can solve this system of equations using substitution or elimination. Let's use substitution:
From Equation 1, we can isolate a: a = 34 - 9d.
Substituting this value for a in Equation 2, we get: 710 = 10 * (2(34 - 9d) + 19d)
Simplifying this equation, we have: 710 = 10 * (68 - 7d).
Dividing both sides of the equation by 10, we get: 71 = 68 - 7d.
Simplifying further, we have: 7d = 68 - 71.
Combining like terms, we have: 7d = -3.
Dividing both sides of the equation by 7, we get: d = -3/7.
Now that we've found the value of the common difference (d), we can substitute it back into Equation 1 to find the first term (a):
34 = a + 9 * (-3/7).
Simplifying this equation, we have: 34 = a - 27/7.
To combine the fractions, we need to have a common denominator. The common denominator is 7, so we can rewrite 34 as 238/7:
238/7 = a - 27/7.
Combining the fractions, we have: 238/7 = (a - 27)/7.
Multiplying both sides of the equation by 7, we get: 238 = a - 27.
Adding 27 to both sides of the equation, we get: 238 + 27 = a.
Simplifying, we have: 265 = a.
Now we have the values of a (the first term) and d (the common difference). We can use the formula for the nth term to find the 25th term (T25):
T25 = a + (25 - 1) * d.
= 265 + 24 * (-3/7).
Simplifying this equation, we have: T25 = 265 - 72/7.
To combine the fractions, we need to have a common denominator. The common denominator is 7, so we can rewrite 265 as 1855/7:
T25 = 1855/7 - 72/7.
Combining the fractions, we have: T25 = (1855 - 72)/7.
Simplifying, we have: T25 = 1783/7.
Therefore, the 25th term of the arithmetic series is 1783/7.