Answer :

Given that a_{2} = 38 & a_{6} = -22

a_{2} = a_{1} + (2-1)d

38 = a_{1} + d ⇒ __1__

a_{6} = a_{1} + (6-1)d

-22 = a_{1} + 5d ⇒ __2__

By subtracting the both equations

38 – (-22) = (a_{1} + d ) - (a_{1} + 5d)

60 = -4d

d = -15

By substituting in __1__ equations we get a_{1}

38 = a_{1} + (-15)

53 = a_{1}

To find a_{3}:

a_{3} = a_{1} + (3-1)d

= 53 + 2×(-15)

= 23

To find a_{4}:

a_{4} = a_{1} + (4-1)d

= 53 + 3×(-15)

= 8

To find a_{5}:

a_{5} = a_{1} + (5-1)d

= 53 + 4×(-15)

= -7

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