# Show that each of the progressions given below is an AP. Find the first term, common difference and next term of each.

Question:

Show that each of the progressions given below is an AP. Find the first term, common difference and next term of each.

(i) $9,15,21,27, \ldots$

(ii) $11,6,1,-4$,

(iii) $-1, \frac{-5}{6}, \frac{-2}{3}, \frac{-1}{2}, \ldots$

(iv) $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots$

(v) $\sqrt{20}, \sqrt{45}, \sqrt{80}, \sqrt{125}, \ldots$

Solution:

(i) The given progression 9, 15, 21, 27, ... .

Clearly, 15 − 9 = 21 − 15 = 27 − 21 = 6 (Constant)

Thus, each term differs from its preceding term by 6. So, the given progression is an AP.

First term = 9

Common difference = 6

Next term of the AP = 27 + 6 = 33

(ii) The given progression 11, 6, 1, −4, ... .

Clearly, 6 − 11 = 1 − 6 = −4 − 1 = −5 (Constant)

Thus, each term differs from its preceding term by −5. So, the given progression is an AP.

First term = 11

Common difference = −5

Next term of the AP = −4 + (−5) = −9

(iii) The given progression $-1, \frac{-5}{6}, \frac{-2}{3}, \frac{-1}{2}, \ldots$

Clearly, $\frac{-5}{6}-(-1)=\frac{-2}{3}-\left(\frac{-5}{6}\right)=\frac{-1}{2}-\left(\frac{-2}{3}\right)=\frac{1}{6}$ (Constant)

Thus, each term differs from its preceding term by $\frac{1}{6}$. So, the given progression is an AP.

First term = −1

Common difference $=\frac{1}{6}$

Next term of the AP $=\frac{-1}{2}+\frac{1}{6}=\frac{-2}{6}=\frac{-1}{3}$

(iv) The given progression $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots$

This sequence can be re-written as $\sqrt{2}, 2 \sqrt{2}, 3 \sqrt{2}, 4 \sqrt{2}, \ldots$

Clearly, $2 \sqrt{2}-\sqrt{2}=3 \sqrt{2}-2 \sqrt{2}=4 \sqrt{2}-3 \sqrt{2}=\sqrt{2}$ (Constant)

Thus, each term differs from its preceding term by $\sqrt{2}$. So, the given progression is an AP.

First term $=\sqrt{2}$

Common difference $=\sqrt{2}$

Next term of the AP $=4 \sqrt{2}+\sqrt{2}=5 \sqrt{2}=\sqrt{50}$

(v) The given progression $\sqrt{20}, \sqrt{45}, \sqrt{80}, \sqrt{125}, \ldots$

This sequence can be re-written as $2 \sqrt{5}, 3 \sqrt{5}, 4 \sqrt{5}, 5 \sqrt{5}, \ldots$

Clearly, $3 \sqrt{5}-2 \sqrt{5}=4 \sqrt{5}-3 \sqrt{5}=5 \sqrt{5}-4 \sqrt{5}=\sqrt{5}$ (Constant)

Thus, each term differs from its preceding term by $\sqrt{5}$. So, the given progression is an AP.

First term $=2 \sqrt{5}=\sqrt{20}$

Common difference $=\sqrt{5}$

Next term of the AP $=5 \sqrt{5}+\sqrt{5}=6 \sqrt{5}=\sqrt{180}$