Question

Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.


NCERT Class X

Mathematics - Mathematics


Chapter _ARITHMETIC PROGRESSIONS

Answer 1

Hey User !!!

Let a be the First term, a3 be the third term, a5 be the 5th term and a7 be the 7th term

a3 = 16

a7 = a5 + 12  ............ (1)

Let the common difference be "d"

Common difference is equal in AP 

So,

a7 = a5 + d + d = a5 + 2d ............(2)

From Equation (1) & (2)

a5 + 12 = a5 + 2d 

2d = 12

d = 6

From Given, we get that

a3 = 16

a3 = a + 2d = 16

a + ( 2 × 6 ) = 16              [ We know that d = 6 ]

a + 12 = 16

a = 4

So first term is 4 .... We can find AP by adding d continuously

So, AP is 4, 10, 16, 22, 28....... 

Hope it helps !!!

Answer 2

Answer:

Explanation:

Solution :-

Let a be the 1st term and d be the common difference of the AP.

We have,

a₃ = 16

a + 2d = 16 ..... (i)

a₇ = a + 6d and a₅ = a + 4d

According to the questions,

⇒ a₇ - a₅ = 12

⇒ a + 6d - (a + 4d) = 12

⇒ a + 6d - a - 4d = 12

⇒ 2d = 12

⇒ d = 12/2

⇒ d = 6

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

a + 2 × 6 = 16

a + 12 = 16

a = 16 - 12 = 4

Here, a = 4, d = 6

Hence, AP is a, a + d, a + 2d, a + 3d +..

4, 4 + 6, 4 + 12, 4 + 18, ...

4, 10, 16, 22..