Question

Help I need it now!! T14= 46, and t20= 100, find t3, t7, and tn

Answer 1

Answer:

$t_3 = -53\\\\\\t_7 = -17\\t_n= -80 +9n$

Step-by-step explanation:

Arithmetic sequence:

       $\sf \boxed{t_n=a+(n-1)d}$

Here a is the first term and d is the common difference.

      $t_{14}= 6$    ⇒  $\sf a + 13 d = 46$  ------------(I)

     $\sf t_{20} = 100$ ⇒   a + 19d = 100 ---------(II)

Subtract equation (I) from (II)

(I1)    a + 19d = 100

(II)    a + 13d   =  46

      -    -            -

               6d = 54

                d = 54 ÷ 6

                 $\sf \boxed{d = 9}$

Substitute d = 9 in equation(I) and find 'a',

    a + 13*9= 46

     a + 117  = 46

              a = 46 - 117

              a = -71

         

$\sf t_3 = -71 + 2*9$

   = -71 + 18

   = -53

          $\sf t_7 = -71 + 6*9$

              = -71 + 54

              = -17

$\sf t_n= -71 + (n-1)*9$

    = -71 + 9*n - 1 *9

    = -71 + 9n - 9

    = -71 - 9 + 9n

     = - 80 + 9n