Given that a2 = 13 & a4 = 3a2 = a1 + (2-1)d13 = a1 + d ⇒ 1a4 = a1 + (4-1)d3 = a1 + 3d ⇒ 2By subtracting the both equations13 – 3 = (a1 + d ) - (a1 + 3d)10 = -2dd = -5By substituting in 1 equations we get a113 = a1 + (-5)18 = a1a3 = a1 + (3-1)d= 18 + 2 × (-5)= 8.

Q. 3 B3.8( 10 Votes )

Find the re

Class 10th AP- Mathematics 6. Progressions

Answer :

Given that a₂ = 13 & a₄ = 3

a₂ = a₁ + (2-1)d

13 = a₁ + d ⇒ 1

a₄ = a₁ + (4-1)d

3 = a₁ + 3d ⇒ 2

By subtracting the both equations

13 – 3 = (a₁ + d ) - (a₁ + 3d)

10 = -2d

d = -5

By substituting in 1 equations we get a₁

13 = a₁ + (-5)

18 = a₁

a₃ = a₁ + (3-1)d

= 18 + 2 × (-5)

= 8.

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