Question

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

Answer 1

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 :

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

Where d exists a common difference

n be the number of terms

a be the first term

Substitute the value of n = 2

$a_{2}=a+(2-1) d=a+d$

Let, the second term exists 12

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

Substitute the value of n = 7

$a_{7}=a+(7-1) d=a+6 d$

Substitute n = 4

$a_{4}=a+(4-1) d=a+3 d$

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:

brainly.com/question/24191546

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