# Show that each of the following sequences is an A.P. Also find the common difference and write 3 more terms in each case.

Question:

Show that each of the following sequences is an A.P. Also find the common difference and write 3 more terms in each case.

(i) 3, −1, −5, −9 ...

(ii) −1, 1/4, 3/2, 11/4, ...

(iii) $\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2}, 7 \sqrt{2}, \ldots$

(iv) 9, 7, 5, 3, ...

Solution:

(i) We have:

$-1-3=-4$

$-5-(-1)=-4$

$-9-(-5)=-4 \ldots$

Thus, the sequence is an A.P. with the common difference being $-4$.

The next three terms are as follows:

$-9-4=-13$

$-13-4=-17$

$-17-4=-21$

(ii) We have:

$1 / 4-(-1)=5 / 4$

$3 / 2-1 / 4=5 / 4$

$11 / 4-3 / 2=5 / 4$

Thus, the sequence is an A.P. with the common difference being $(5 / 4)$.

The next three terms are as follows:

$11 / 4+5 / 4=16 / 4=4$

$16 / 4+5 / 4=21 / 4$

$21 / 4+5 / 4=26 / 4$

(iii) We have:

$3 \sqrt{2}-\sqrt{2}=2 \sqrt{2}$

$5 \sqrt{2}-3 \sqrt{2}=2 \sqrt{2}$

$7 \sqrt{2}-5 \sqrt{2}=2 \sqrt{2}$

Thus, the sequence is an A.P. with the common difference being $(2 \sqrt{2})$.

The next three terms are as follows:

$7 \sqrt{2}+2 \sqrt{2}=9 \sqrt{2}$

$9 \sqrt{2}+2 \sqrt{2}=11 \sqrt{2}$

$11 \sqrt{2}+2 \sqrt{2}=13 \sqrt{2}$

(vi) We have:

$7-9=-2$

$5-7=-2$

$3-5=-2$

Thus, the sequence is an A.P. with the common difference being $(-2)$.

The next three terms are as follows:

$3-2=1$

$1-2=-1$

$-1-2=-3$