Question

2. Find the 20th term of an arithmetic sequence if its 6th term is 14 and 14th term is 6.

Answer 1

Answer:

$\sf t_{20}= 0$

Step-by-step explanation:

Arithmetic sequence:

      $\sf \boxed{\bf n^{th} \ term = a + (n-1)d}\\\\\text{Here, a is the first term ; d is the common difference }$

6th term is 14 ⇒ $\sf t_6 = 14$

                a + (6 - 1)d = 14

                    a  +  5d = 14  --------------(I)

14th term is 6 ⇒$\sf t_{14} = 6$

             a + (14-1)d = 6

                  a + 13d = 6 ----------------(II)

Subtract equation (II) from equation(I)

        (I)          a + 5d = 14

        (II)         a + 13d = 6

                    -    -          -

                            -8d = 8

                               d  = 8 ÷(-8)      

                              $\sf \boxed{\bf d= (-1)}$

Plugin d = -1 in equation (I)

a + 5(-1) = 14

      a -5  = 14

             a = 14 + 5

             $\sf \boxed{\bf a = 19}$  

20th term:

 $\sf t_{20}= 19 + 19*(-1)$

       = 19 - 19

   $\sf \boxed{\bf t_{20} = 0}$

Answer 2

Answer:

0

Step-by-step explanation:

The number of terms of an Arithmetic progressions has the formular.

Tn = a + ( n - 1 ) d

From the question,

6th term = 14

14th term = 6

Therefore,

a + 5d = 14 -----------(1)

a + 13d = 6 ----------(2)

subtracting

-8d = 8

dividing bothsides by -8

$\frac{ - 8d}{ - 8} = \frac{8}{ - 8} \\ d = - 1$

Therefore,

common difference= -1

substituting the value of d into equation (1)

a + 5 ( -1) = 14

a - 5 = 14

a = 14 + 5 = 19

First term = 19

For the 20th term

T 20 = a + 19d

19 + 19 ( -1 )

19-19 = 0

Therefore,

20th term = 0