Question

10th term of an A.P is 66 and if it's common difference is 7,find position of 150 in this sequence​

Answer 1

a10=66=a+9d

d=7

an=150

so let us take

a10=a+9d

66=a+9×7

66-63=a

3=a

now we take

an=a +(n-1)d

150=3+(n-1)7

150-3=(n-1)7

143=(n-1)7

143/7 +1=n

143/7+7/7

150/7=n

Answer 2

10th term =66
=A + 9D =66,where a is first term and d is common difference
Given d=7
From 1st equation
A= 66-(7X9)
66-63=9
To find Position of 150
Take An=150
An=A+(n-1)d
150=9+7n-7
=150=7n+2
=7n=150-2
=n=148/7
n=you will get a decimal value so question is wrong.It cant be 150