Question

if t2=-12, t8=6, find tn and t6​

Answer 1

Solution:

t2=-12=a+d equation 1

t8=6=a+7d equation 2

subracting equation 1 and 2

a+d-(a+7d)=(-12)-6

-6d=-18

d=3

substituting value of d in equation 1

a+3=-12

a=-15

tn=a+(n-1)d

=-15+3n-3

=-18+3n

t6=-18+3(6)

=-18+18

=0

Answer 2

Answer:

Required value of $t_n$ is (3n-18) and value of $t_6$ is 0

Step-by-step explanation:

Given,

$t_2 =(-12) \\ t_8 = 6$

We know,

$t_n = a + (n - 1)d$

Where a is the first term and d is common difference of a series and n is number of term.

According to question,

$t_2 = a + (2 - 1)d \\ (-12) = a + d \\ a + d = -12....(1)$

and

$t_8 = a + (8 - 1)d = 6 \\ a + 7d = 6....(2)$

We are subtracting equation (2) from equation (1),

$a + 7d - (a + d) = 6 + 12 \\ a + 7d - a - d = - 18 \\ 6d = + 18 \\ d = \frac{ 18}{6} \\ d = 3$

From equation (1),

$a +3 = -12 \\ a = -12 -3 \\ a = -15$

Now,

$t_6 = a + (n - 1)d \\ = -15 + (6 - 1)\times3 \\ = -15+15 \\ = 0$

and

$t_n = a + (n - 1)d \\ = -15+ (n - 1) \times ( 3) \\ = +3n -18$

This is a problem of Arithmetic progression.

Know more about Arithmetic progression:

1) https://brainly.in/question/4219484

2) https://brainly.in/question/2768711

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