Question

determine the AP whose third term is 16 and 7th term exceeds the 5th term by 12​

Answer 1

Solution:

Given:

⇒ a₃ = 16

⇒ a₇ - a₅ = 12

To Find:

⇒ A.P

Formula used:

⇒ an = a + (n - 1)d

So, As we know, a₃ = a + 2d

⇒ a + 2d = 16     ...........(1)

Similarly,

⇒ a₇ = a + 6d

⇒ a₅ = a + 4d

It is given that 7th term exceeds the 5th term by 12​.

⇒ a₇ - a₅ = 12

⇒ (a + 6d) - (a + 4d) = 12

⇒ a + 6d - a - 4d = 12

⇒ 2d = 12

⇒ d = 12/2

⇒ d = 6

Now, put the value of d in Equation (1), we get

⇒ a + 2d = 16  

⇒ a + 2(6) = 16

⇒ a + 12 = 16

⇒ a = 16 - 12

⇒ a = 4

Hence,

⇒ First term (a₁) = 4

⇒ Second term (a₂) = a + d = 4 + 6 = 10

⇒ Third term (a₃) = a + 2d = 4 + 2(6) = 16

⇒ Fourth term (a₄) = a + 3d = 4 + 3(6) = 22

So, the A.P is 4, 10, 16, 20,.........

Answer 2

SOLUTION:-

Given:

A.P.whose third term is 16 & 7th term exceeds the 5th term by 12.

To find:

Arithmetic Progression.

Explanation:

Formula of the A.P.

${}^{a} n = a + (n - 1)d$

• First term= a
• Common difference= d
• Number term= n

${}^{a} 3 = a + (3 - 1)d \\ \\ {}^{a}3 = a + 2d \\ \\ 16 = a + 2d \\ \\ a + 2d = 16 ...........(1)$

&

${}^{a} 7 = a + (7 - 1)d \\ \\ {}^{a}7 = a + 6d.............(2)$

Similarly,

${}^{a} 5 = a + (5 - 1)d \\ \\ {}^{a}5 = a + 4d..........(3)$

Now,

7th term exceeds the 5th term by 12:

${}^{a} 7 - {}^{a} 5 = 12$

=) (a+6d) - (a+4d)=12

=) a+6d - a -4d= 12

=) a-a +6d -4d= 12

=) 2d = 12

=) d= 12/2

=) d= 6

Putting the value of d in equation (1), we get;

=) a+ 2×6= 16

=) a+ 12= 16

=) a= 16 -12

=) a= 4

Hence,

First term of A.P., a= 4

Second term of A.P.= first term+common difference

Second term= 4 + 6= 10

Third term of A.P.= 10 +6= 16

Thus,

The A.P. is 4,10,16, 20.....