Question

1)State whether the following sequence is an A.P or not

125, 132, 137, 142,…….​

Answer 1

Answer:

since the difference between consecutive terms (132-125=7 and 137-132=5) is not equal it is not an AP

Step-by-step explanation:

a1=125

a2=132

a3=137

a2-a1=7

a3-a2=5

5≠7

the given series is not an AP