Answer :

Given that a_{2} = 13 & a_{4} = 3

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

13 = a_{1} + d ⇒ __1__

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

3 = a_{1} + 3d ⇒ __2__

By subtracting the both equations

13 – 3 = (a_{1} + d ) - (a_{1} + 3d)

10 = -2d

d = -5

By substituting in __1__ equations we get a_{1}

13 = a_{1} + (-5)

18 = a_{1}

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

= 18 + 2 × (-5)

= 8.

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