Question

determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12​

Answer 1

Answer:

A3 = 16

A7 =A5+12

Now An = A + (n-1) d

A3= A+ 2d

A5= A + 4d

A7=A +6d

Now A + 6d = A + 4d +12

Hence , A + 6d -A -4d=12

2d=12

d=6

Now 16= A +2d

16=A +12

A=4

Ap= 4,10,16

Answer 2

t3=16 or a+2d=16.........(1)

and t7-t5=12

or. a+6d-(a+4d)=12

or 2d =12

or d= 6

now put the value of d in equation (1)

a=4

isliye. a1= 4,a2=a+d=10,a3=a+2d=16,a4=a+3d=22...........as shown on....