Question

Find arithmetic progression whose third term is 16 and the seventh term exceeds the 5th term by 12

Answer 1

Answer:

Step-by-step explanation:

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

$\mathfrak{\huge{Answer:}}$

If we let the first term of the Arithmetic Progression to be = a

The common difference to be = d

Then :-

$\mathbb{GIVEN}$

$\sf{a_{3} = 16}$

$\sf{a_{7} = a_{5} + 12}$

$\mathbb{TO\:FIND}$

The arithmetic progression

$\mathbb{METHOD}$

$\sf{a_{5} = a + (n - 1)d}$

Solve this formed equation further

=》 $\sf{a_{5} = a + 4d}$ ....(1)

Similarly :-

$\sf{a_{7} = a + 6d}$ ....(2)

We're already given with a relation between the fifth and the seventh term. Using it, we get :-

=》 $\sf{a + 4d + 12 = a + 6d}$

Solve this formed equation further

=》 $\sf{2d = 12}$

=》 d = 6

Keep this value in $\sf{a_{3} = 16}$ and get the value of a :-

=》 $\sf{a_{3} = 16 = a + 2d}$

Solve this formed equation further

=》 $\sf{a + 12 = 16}$

=》 a = 4

Arithmetic Progression = $\huge{\tt{4,10,16,22,28...}}$