Question

show that the sequence described by an=1/3n+1/6 is an a.p

Answer 1

Step-by-step explanation:

We know that in an A.P.,

$t_{n} - t_{n - 1} = d$

Where d is the common difference.

Tn=1/3n+1/6

=(2+n)/6n

Tn-1=1/3(n-1)+1/6

=(2-17n)/6n

Tn-Tn-1=(2+n-2+17n)/6n

=3

d=3, so d is a constant and independent on n.

Hence, proved

Answer 2

Since $d_{1}= d_{2}$, the sequence is an A.P

Explanation:

Given: A sequence described by $a_{n} =\frac{1}{3}n+\frac{1}{6}$

To prove: The sequence described is an A.P

Proof: We have, $a_{n} =\frac{1}{3}n+\frac{1}{6}$

Put $n=1$

$a_{1} =\frac{1}{3}+\frac{1}{6}=\frac{1}{2}$

So, $a_{1} =\frac{1}{2}$

Now, put $n=2$

$a_{2} =\frac{1}{3}(2)+\frac{1}{6}=\frac{5}{6}$

So, $a_{2} =\frac{5}{6}$

Now, put $n=3$

$a_{3} =\frac{1}{3}(3)+\frac{1}{6}=\frac{7}{6}$

So, $a_{3} =\frac{7}{6}$

Now, put $n=4$

$a_{4} =\frac{1}{3}(4)+\frac{1}{6}=\frac{3}{2}$

So, $a_{4} =\frac{3}{2}$

Hence, we get the sequence

$\frac{1}{2}, \frac{5}{6}, \frac{7}{6}, \frac{3}{2},...$

Now,

$d_{1} =\frac{5}{6}-\frac{1}{2}=\frac{1}{3}$

$d_{2} =\frac{7}{6}-\frac{5}{6}=\frac{1}{3}$

Clearly, $d_{1}= d_{2}$

Hence, the sequence is an AP.

Learn more:

Sequence and AP

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