Question

Find the ap whose second term is 12th and 7th term exceeds the 4th term by 15

Answer 1

a2=12 a+d=12 eq 1

a7-a4=15

a+6d -( a+3d)/15

a+6d-a-3d=15

3d=15 (a-a=0)

d=15/3

d= 5

so put in eq 1

a+d=12

a+5=12

a=12-5

a=7

so ap is

7,12,17,...... so on

I hop it help u

thank

Answer 2

Given :

Second term is 12th and 7th term exceeds the 4th term by 15

To Find :

Find AP

Solution:

Formula of nth term :

$a_n=a+(n-1)d$

Where d is common difference

n = no. of terms

a = first term

Substitute n = 2

$a_2=a+(2-1)d=a+d$

We are given that second term is 12

So,a+d=12 ---- 1

Substitute n = 7

$a_7=a+(7-1)d=a+6d$

Substitute n = 4

$a_4=a+(4-1)d=a+3d$

We are given that 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=7

a=7

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

So, AP : 7,12,17,....