Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Discover a wealth of knowledge from experts across different disciplines on our comprehensive Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

Find the A.P whose second term is 12 and 7th term exceeds the 4th by 15

Sagot :

The 2nd term exists 12 and 7th term exceeds the 4th by 15 exists AP : 7,12,17,....

How to estimate the arithmetic progression?

Given : The second term exists 12th and the 7th term exceeds the 4th term by 15

The formula of the nth term :

[tex]$a_{n}=a+(n-1) d[/tex]

Where d exists a common difference

n be the number of terms

a be the first term

Substitute the value of n = 2

[tex]$a_{2}=a+(2-1) d=a+d[/tex]

Let, the second term exists 12

So, a + d = 12 ..........(1)

Substitute the value of n = 7

[tex]$a_{7}=a+(7-1) d=a+6 d[/tex]

Substitute n = 4

[tex]$a_{4}=a+(4-1) d=a+3 d[/tex]

We are given that the 7th term exceeds the 4th term by 15

So, a+6d-a-3d = 15

3d = 15

d = 5

Substitute the value of d in (1)

So, a+5 = 12

a = 12-5

a = 7

AP : a,a+d,a+2d,.... = 5, 7+5, 7+2(5),....=7,12,17,....

Therefore, AP : 7,12,17,....

To learn more about arithmetic progression refer to:

https://brainly.com/question/24191546

#SPJ9