SOLUTION: the 10th term of an arithmetic seq. is 25 and the sum of the 1st 20 terms is 530. find the 17th term of the seq.

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Question 710624: the 10th term of an arithmetic seq. is 25 and the sum of the 1st 20 terms is 530. find the 17th term of the seq.

Found 2 solutions by srinivas.g, stanbon:
Answer by srinivas.g(540) (Show Source):
You can put this solution on YOUR website!
10 th term =25
In A.p, formula for n th term = a +(n-1)d
where a = initial or starting term
d= common difference
since 10 th term = 25
a+(10-1)d =25
a+9d =25............. ...(1)
sum of n terms = (first term + last term ) * no of terms / 2

here last term = 20 th term = a+(20-1)d
=a+19 d
sum = (a+(a+19d)) *20 /2
530 = ( 2a+19d ) *10
divide with 10 on both sides
530/10 = (2a+19d) * 10/10
53 = 2a+19d *1
so 2a+19d = 53 ...................(2)
so we need to solve equations (1) and (2) to get 'a ' and ' d'
multiply eq (1) with 2
2*(a+9d) = 2(25)
2a+18d =50 ...............(3)
subtract eq (3) from eq(2)
2a + 19d -(2a+18d) = 53-50
d =3
put d= 3 in eq(1) a+9(3) =25
a= -2
so 17 th term = a+(17-1)d
= -2 +16*3
=-2 +48
=46

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
the 10th term of an arithmetic seq. is 25 and the sum of the 1st 20 terms is 530. find the 17th term of the seq.
---
Equations:
a(10) = a + 9d = 25
a(20) = a + 19d
----
S(20) = (20/2)(a + (a+19d)) = 530
10(2a+19d) = 530
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Equations:
a + 9d = 25
2a+19d = 53
-----
Modify:
2a + 18d = 50
2a + 19d = 53
---
Subtract to get:
d = 3 (the common difference)
Solve for "a":
a + 9d = 25
a + 27 = 25
a = -2
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Seventeenth term:
a(17) = a + 16d
a(17) = -2 + 16*3 = 46
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Cheers,
Stan H.
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