Question

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 is 28 and 8th term of second ap is 29 , then find both .

Answer 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. . . . .

Answer 2

Answer:

Given :-

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 is 28 and 8th term of second ap is 29

To Find :-

Both AP

Solution :-

Let the first AP be a and second AP be a'

Now

$\bf \: a_{n} = a + (n - 1)d$

First term is 3 more than Second so,

a = a' + 3

For the first AP it is given that the seventh term is 28

$\sf a_7 = a(n - 1)d$

$\sf 28 = a + (7 - 1)d$

$\sf \: 28 = a + (6)d$

$\sf \: 28 = a + 6d$

$\sf 28 = a' + 3 + 6d$

$\sf 28 - 3 = a' + 6d$

$\sf 25 = a' + 6d$

For second AP

$\sf a_8 = a+(n-1)d$

$\sf29 = a + (8 - 1)d$

$\sf \: 29 = a + (7)d$

$\sf \: 29 = a + 7d$

Finding common Difference

$\sf \: 25 - 6d + 7d = 29$

$\sf \: 25 - d = 29$

$\sf \: - d = 25 - 29$

$\sf \: - d = - 4$

$\sf \: d = 4$

For the second term

a' = 25 - 6 × 4

a' = 25 - 24

a' = 1

AP = 1,5,9,

For the first term

a = 1 + 3

a = 4

AP = 4,8,12