Question

Which of the following sequences are A.P. ? If they are A.P. find the common difference .
(1) 2, 4, 6, 8,....
(2) 2, $\frac{5}{2}$, 3, $\frac{7}{3}$ ,....
(3) –10, –6, –2, 2...
(4) 0.3, 0.33, 0.333,...
(5) 0, –4, –8, –12,...
(6) $-\frac{1}{5},-\frac{{\displaystyle 1}}{{\displaystyle 5}},-\frac{{\displaystyle 1}}{{\displaystyle 5}},...$
(7) 3, $3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2}$
(8) 127, 132, 137,...

Answer 1

(1) 2, 4, 6, 8,....

The given sequence is an A.P.
Common difference = Second term − First term
= 4 − 2
= 2

(2) 2, $\frac{5}{2}$, 3, $\frac{7}{3}$ ,....

The given sequence is not an A.P.

(3) –10, –6, –2, 2...

The given sequence is an A.P.
Common difference = Second term − First term
= (−6) − (−10)
= 4

(4) 0.3, 0.33, 0.333,...

The given sequence is not an A.P.

(5) 0, –4, –8, –12,...

The given sequence is an A.P.
Common difference = Second term − First term
= (−4) − (0)
= −4

(6) $-\frac{1}{5},-\frac{{\displaystyle 1}}{{\displaystyle 5}},-\frac{{\displaystyle 1}}{{\displaystyle 5}},...$

The given sequence is an A.P.
Common difference = Second term − First term
= (−$\frac{1}{5}$) − (−$\frac{1}{5}$)
= 0

(7) 3, $3+\sqrt{2},3+2\sqrt{2},3+3\sqrt{2}$

The given sequence is an A.P.
Common difference = Second term − First term
= $3+\sqrt{2}-3$
= $\sqrt{2}$

(8) 127, 132, 137,...

The given sequence is an A.P.
Common difference = Second term − First term
= (132) − (127)
= 5