Question

25 POINTS. PLEASE HELP

Answer 1

t1 = a1 = -13/2

t2 = -10/2

Explanation:

Sn = n/2 [2a1 + (n - 1)d] is the formula for the sum of an arithmetic series

an = a1+ (n-1)d is the formula for the nth term

d= common difference and a1 is the initial term

a12 = 10

10 = a1 + (12-1)d

10 = a1 +11d

21 = 12/2 (2a1 + (12-1)d)

21 = 6 (2a1 +11d)

Distribute

21 = 12a1 +66d

We have 2 equations and 2 unknowns

10 = a1 +11d

21 = 12a1 +66d

Multiply the first equation by -12

-12 *10 = -12(a1 +11d)

-120 =- 12a1 - 132d

Add this to the second equation

-120 =- 12a1 - 132d

21 = 12a1 +66d

-----------------------

-99 = -66d

Divide by -66

-99/-66 = -66d/-66

3/2 =d

Now we can find a1

10 = a1 +11d

10 = a1 + 11 (3/2)

10 = a1 +33/2

Subtract 33/2

10-33/2 = a1-33/2+33/2

20/2 -33/2 = a1

-13/2 =a1

t1 = a1 = -13/2

t2 = a1+ (2-1)3/2

= -13/2 + 3/2

= -10/2

Answer 2

The formula of a sum of terms of an arithmetic sequence:

`S_n=(t_1+t_n)/(2)\cdot n`

We have

`S_(12)=21,\ t_(12)=10,\ n=12`

Substitute:

`21=(t_1+10)/(2)\cdot12`

`21=(t_1+10)\cdot6` use the distributive property: a(b + c) = ab + ac

`21=6t_1+60` subtract 60 from both sides

`-39=6t_1` divide both sides by 6

`t_1=-(39)/(6)\\\\t_1=-(13)/(2)`

The explicit form of an arithmetic sequence:

`t_n=t_1+(n-1)d`

d - common difference

`d=t_(n+1)-t_n\to 11d=t_(12)-t_1`

Substitute:

`11d=10-\left(-(13)/(2)\right)`

`11d=(20)/(2)+(13)/(2)`

`11d=(33)/(2)` divide both sides by 11

`d=(3)/(2)`

`t_2=t_1+d\to t_2=-(13)/(2)+(3)/(2)=-(10)/(2)=-5`

Answer:

`t_1=-(13)/(2)\ and\ t_2=-5`