Question

in the following APs,find the missing terms in the boxes
1). 5_,_,9/2
2).-4,_,_,_,_,6
3)._,38,_,_,_,-22​

Answer 1

Answer:

see explanation

Step-by-step explanation:

The n th term of an AP is

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

(1)

Given a₁ = 5 and a₄ = $\frac{9}{2}$ , then

5 + 3d = $\frac{9}{2}$ ( subtract 5 from both sides )

3d = - $\frac{1}{2}$ ( divide both sides by 3 )

d = - $\frac{1}{6}$

add - $\frac{1}{6}$ to each term

5,  $\frac{29}{6}$,  $\frac{14}{3}$,  $\frac{9}{2}$

(2)

Given a₁ = - 4 and a₆ = 6, then

- 4 + 5d = 6 ( add 4 to both sides )

5d = 10 ( divide both sides by 5 )

d = 2

Add 2 to each term

- 4,  -2,  0,  2,  4,  6

(3)

Given a₂ = 38 and a₆ = - 22, then

a₁ + d = 38 → (1)

a₁ + 5d = - 22 → (2)

Subtract (1) from (2) term by term

4d = - 60 ( divide both sides by 4 )

d = - 15

Substitute d = - 15 into (1)

a₁ - 15  = 38 ( add 15 to both sides )

a₁ = 53

Thus

53,  38,  8,  - 7,  - 22