the 10th term is 34 of arithmetic series and the sum of the 20th term is 710.what is the 25th term?
the difference between T10 and T20 is 10d. So, 10d=710-34=676
d=67.6
T25=T20+5d = 710+338=1048
the answer is t25=79
and it sum of 20th term that is S20=710.
you cannot have the "sum of the 20th term".
If you mean the sum of the 1st 20 terms, then
T10 = a+9d=34
S20 = 10(2a+19d) = 710
so, a=7, d=3
T25 = 7+24*3 = 79
To find the 25th term of an arithmetic series, we first need to find the common difference (d) of the series.
Given that the 10th term (a10) is 34, we can use the formula for the nth term of an arithmetic series:
an = a1 + (n - 1)d
where an represents the nth term, a1 is the first term, n is the position of the term, and d is the common difference.
Using the formula, we can substitute the given values:
34 = a1 + (10 - 1)d
Now, we need to find the sum (S20) of the 20th term. The formula for the sum of an arithmetic series is:
Sn = (n/2)(a1 + an)
Substituting the given values:
710 = (20/2)(a1 + a20)
We have two equations with two unknowns:
34 = a1 + 9d (equation 1)
710 = 10(a1 + a20) (equation 2)
We can solve these equations simultaneously to find the values of a1 and d.
From equation 1, we can isolate a1:
a1 = 34 - 9d
Now substitute this value into equation 2:
710 = 10((34 - 9d) + a20)
Simplify the equation:
710 = 340 - 90d + 10a20
Rearrange the equation:
90d - 10a20 = -30
Now we have a system of equations:
34 - 9d = a1 (equation 1)
90d - 10a20 = -30 (equation 2)
To solve this system, we can start by solving equation 2 for a20:
a20 = (90d + 30)/10
= 9d + 3
Substitute this value of a20 into equation 1:
34 - 9d = a1
34 - 9d = 9d + 3
Solve for d:
34 - 3 = 9d + 9d
31 = 18d
d = 31/18
Substitute the value of d back into equation 1 to find a1:
a1 = 34 - 9(31/18)
= 34 - (279/18)
= 340/18 - 279/18
= 61/18
Now that we have the values of a1 and d, we can find the 25th term (a25) of the arithmetic series using the formula:
a25 = a1 + (25 - 1)d
Substituting the values:
a25 = (61/18) + (24)(31/18)
Simplify the expression:
a25 = (61 + 744)/18 = 805/18
Therefore, the 25th term of the given arithmetic series is 805/18.