Thursday

October 30, 2014

October 30, 2014

Posted by **Nancy** on Saturday, February 13, 2010 at 10:55pm.

- Math/Precal -
**MathMate**, Saturday, February 13, 2010 at 11:23pmFor an arithmetic sequence, the nth term is given by:

T(n)=T(0)+kn,

where k is a constant.

Knowing T(7)=21, and T(10)=126

we find k using

T(10)-T(7) = 126-21

T(0)+10k - (T(0)+7k) = 105

3k = 105

k=35

From the value of k, we find T(0):

T(n)=T(0)+35n

T(7)=T(0)+35(7) = 21

T(0)=21-245=-224

T(n) = -224 + 35n

- Math/Precal -
**Reiny**, Saturday, February 13, 2010 at 11:35pm7th term is 21 ----> a+6d = 21

10th term is 126 ---> a + 9d = 126

subtract them:

3d = 105

d = 35

back in a+6d = 21 -----> a = -189

first term is -189

check:

t_{7}= -189 + 6(35) = 21

t_{10}= -189 + 9(35) = 126

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