10th term of an A.P is 66 and if it's common difference is 7,find position of 150 in this sequence
Answers
Answered by
0
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
Answered by
1
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
=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
Similar questions