Question

An ap is given by k, 2k/3, k/3, 0, . . .. (i) find the sixth term. (ii) find the n th term. (iii) if the 20th term is equal to 15, find k.

Answer 1

AP: k,2k/3,k/3,0.....................

Here;

a = k

d = a₂ - a₁ = 2k/3-k

d = -k/3

1) a₆ = a+5d

= k + 5*-k/3

= k+-5k/3

a₆ = -2k/3

2) an = k + (n-1)-k/3

= k -kn/3+k/3

= k/3(3-n+1)

= k/3(4-n)

3) a₂₀ = a₁₅

a+19d = 15

k+19*-k/3 = 15

k-19k/3 = 15

3k-19k/3 = 15

-16k = 45

k = -45/16

Answer 2

The answers are as follows:

(i) $\frac{-2k}{3}$

(ii) $\frac{k}{3} (4 - n)$

(iii) $\frac{-45}{16\\}$

Given,

AP = k, 2k/3, k/3, 0....

To Find,

The sixth term, the nth term, and k.

Solution,

First-term (a) = k

Common difference (d) = 2k/3 - k

                                       = - k/3

(i) T₆ = a + (n-1)d

        = k + 5(-k/3)

        = k - 5k/3

∴ T₆  = -2k/3

(ii) $T_{n}$ = a + (n-1)d

         = k + (n-1)(-k/3)

         = k + k/3 - kn/3

         = 4k/3 - kn/3

∴  $T_{n}$   = k/3 (4 - n)

(iii)  $T_{20}$  = 15

     a + 19d = 15

     k + 19(-k/3) = 15

     -16k/3 = 15

      -16k =45

∴ k = -45/16