Question

In an AP , Tn=3n-1, the common difference is

Answer 1

Answer:

Give, nth term: T(n) = 3n – 1

So our (n+1)th will be, T(n+1) = 3(n+1) - 1 = 3n + 2

Now the common difference, d = T(n+1) – T(n) = 3n + 2 - 3n + 1 = 3.

Answer 2

Given:

n an AP $\[{T_n} = 3n - 1\]$, the common difference is-

Solution:

Common difference (d)$\[ = {T_n} - {T_{n - 1}}\]$

Find the first term by putting $n=1$,

$\[\begin{array}{l} \Rightarrow {T_1} = 3 \times 1 - 1\\ \Rightarrow {T_1} = 3 - 1\\ \Rightarrow {T_1} = 2\end{array}\]$

Find the second term by putting $n=2$,

$\[\begin{array}{l} \Rightarrow {T_2} = 3 \times 2 - 1\\ \Rightarrow {T_2} = 6 - 1\\ \Rightarrow {T_2} = 5\end{array}\]$

Therefore,

$\[\begin{array}{l} \Rightarrow d = {T_2} - {T_1}\\ \Rightarrow d = 5 - 2\\ \Rightarrow d = 3\end{array}\]$

Hence, the common difference is 3.