Show that the sequence 9, 12, 15,18 is an AP. Find it's 16th term and the general term ?
Answers
Answered by
134
Hi,
Here is your answer,
Given,
a(First term) = 9, d(common difference) 12 - 9 = 3
Formula:- an = a+(n-1)d
an = 9+(n-1)3
an = 9 + 3n - 3
an = 6 + 3n
Now, 16th term of an A.P,
a₁₆ = 6 + 3 × 16
a₁₆ = 6 + 48
a₁₆ = 54
Therefore the 16th term of an A.P is 54.
Hope it helps you !
Here is your answer,
Given,
a(First term) = 9, d(common difference) 12 - 9 = 3
Formula:- an = a+(n-1)d
an = 9+(n-1)3
an = 9 + 3n - 3
an = 6 + 3n
Now, 16th term of an A.P,
a₁₆ = 6 + 3 × 16
a₁₆ = 6 + 48
a₁₆ = 54
Therefore the 16th term of an A.P is 54.
Hope it helps you !
Answered by
3
Given,
A sequence 9, 12, 15, 18
To Find,
Whether the sequence is A.P or not and the 16th and general term.
Solution,
The given sequence is 9, 12, 15, 18
d = 12-9 = 15-12 = 3
Since the common difference between each term is the same. So the given sequence is an A.P.
So,
a₁₆ = a+15d
a₁₆ = 9+15*3
a₁₆ = 9+45 = 54
Now,
aₙ = a+(n-1)d
aₙ = 9+(n-1)3
aₙ = 9+3n-3
aₙ = 6+3n
Hence, the 16th term of the A.P is 54 and the nth term is 6+3n.
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