Question

determine the A.P whose third term is 16 and 7th term exceeds the 5th term by 12.​

Answer 1

Answer:

Hi mate

Here is your answer..

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

So, AP is 4, 10, 16, 22, 28....... 

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 Helpful ✌️❤️

Answer 2

GIVEN:

• Third term of an A.P = 16
• 7th term exceeds the 5th term by 12.

TO FIND:

• What is the arithmetic sequence ?

SOLUTION:

We have given that, the third term is 16 and 7th term exceeds the 5th term by 12.

• a3 = 16
• a7 = a5 + 12

CASE:- 1)

◇ The third term of an A.P is 16.

➸ a + 2d = 16....❶

CASE:- 2)

◇ 7th term exceeds the 5th term by 12.

➸ a + 6d = a + 4d + 12

➸ a +6d –a –4d = 12

➸ 2d = 12

➸ $\sf{ d = \cancel\dfrac{12}{2}}$

 ❮ d = 6 ❯ 

Put the value of d in equation 1)

➸ a + 2 $\times$ 6 = 16

➸ a + 12 = 16

➸ a = 16 –12

 ❮ a = 4 ❯ 

The Arithmetic sequence is:-

❐ a = 4

❐ a + d = 4 + 6 = 10

❐ a + 2d = 4 +2(6) = 16

❐ a + 3d = 4 +3(6) = 22

❝ Hence, the arithmetic sequence is 4, 10, 16, 22 ❞

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