Question

Find the arithmetic progression whose third term is 16 and 7th term exceeds its fifth term by 12

Answer 1

Answer:

4,10,16....

Step-by-step explanation:

• a3= 16

a+2d=16    ----1

• a7-a5= 12

a+6d- {a+4d}=12

a+6d-a-4d=12

2d=12

d=12/2

d=6.....substitute in 1

NOW

a+2{6}=16

a+12=16

a=16-12

a=4

so the general form of AP IS  a , a+d,a+2d

therefore the AP is 4, 4+6 , 4+ 2{6}

                                4,10,16