The common difference of two ap are equal. The 1st term of an ap is 3 more than the 1st term of second ap. if the 7th term of ap i

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1. Answer:

   First AP: 4, 8, 12, 16. . . . .

   Second AP: 1, 5, 9, 13. . . . .

   Step-by-step explanation:

   Given that:

   The common difference of two AP are equal.

   To Find:

   Both the AP.

   We know that:

   aₙ = a + (n – 1)d

   Where,

   aₙ = nth term

   a = First term

   n = Number of terms

   d = Common difference

   Let us assume:

   Common difference be d.

   First term of first AP be a.

   First term of second AP be A.

   The 1st term of an AP is 3 more than the 1st term of second AP:

   ⇢ a = A + 3 _____(i)

   The 7th term of AP is 28:

   ⇢ a₇ = 28

   ⇢ a + (7 – 1)d = 28

   ⇢ a + 6d = 28

   Substituting the value of a from eqⁿ(i).

   ⇢ A + 3 + 6d = 28

   ⇢ A = 28 – 3 – 6d

   ⇢ A = 25 – 6d _____(ii)

   8th term of second AP is 29:

   ⇢ A₈ = 29

   ⇢ A + (8 – 1)d = 29

   ⇢ A + 7d = 29

   Substituting the value of A from eqⁿ(ii).

   ⇢ 25 – 6d + 7d = 29

   ⇢ 25 + d = 29

   ⇢ d = 29 – 25

   ⇢ d = 4

   In equation (ii).

   ⇢ A = 25 – 6d

   Putting the value of d.

   ⇢ A = 25 – 6(4)

   ⇢ A = 25 – 24

   ⇢ A = 1

   In equation (i)

   ⇢ a = A + 3

   Putting the value of A.

   ⇢ a = 1 + 3

   ⇢ a = 4

   Finding first AP:

   Common difference = d = 4

   First term of first AP = a = 4

   Second term = a + d = 4 + 4 = 8

   Third term = a + 2d = 4 + 2(4) = 12

   Fourth term = a + 3d = 4 + 3(4) = 16

   ∴ First AP: 4, 8, 12, 16. . . . .

   Finding second AP:

   Common difference = d = 4

   First term of second AP = 1

   Second term = a + d = 1 + 4 = 5

   Third term = a + 2d = 1 + 2(4) = 9

   Fourth term = a + 3d = 1 + 3(4) = 13

   ∴ Second AP: 1, 5, 9, 13. . . . .

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