Question

he 4th term of AP is thrice the first and the 7th term exceeds twice the third term by 11. Find AP​

Answer 1

Answer:

Check the Attachment

Hope it helps

Answer 2

Step-by-step explanation:

4th term=3(first term)

a+3d=3a

3d=3a-a=2a

So d =2a/3.....eq.1

7th term=2(third term)+11

a+6d=2(a+2d)+11

a+6d=2a+4d+11

6d-4d=2a-a+11

2d=a+11

d=(a+11)/2......eq.2

From 1 and 2

2a/3=(a+11)/2

2(2a)=3(a+11)

4a=3a+33

4a-3a=33

a=33

Put this value in equation 1

d=(2×33)/3 = 22

So, first term =a=33

Second term =a+2d=33+2(22)

=33+44=77

Third term =a+2d+d=77+22=99

So AP,

33,77,99...........