Question

Write the sequence with $n$th term. Show that all the sequences form A.P $a_n=9-5n$

Answer 1

Given, the nth term of the sequence is

$a_n=9-5n...................eq\left(1\right)$

1) for finding the first term of sequence $\left(a_1\right)$ put n=1 in eq(1)

$a_1=9-5\times 1$

$⟹a_1=9-5$

$⟹a_1=4$

2) for finding the second term of sequence $\left(a_2\right)$put n=2 in eq(1)

$a_2=9-5\times 2$

$⟹a_2=9-10$

$⟹a_2=-1$

3) for finding the third term of sequence $\left(a_3\right)$ put n=3 in eq(1)

$a_3=9-5\times 3$

$⟹a_3=9-15$

$⟹a_3=-6$

4) for finding the fourth term of sequence $\left(a_4\right)$ put n=4 in eq(1)

$a_4=9-5\times 4$

$⟹a_4=9-20$

$⟹a_4=-11$

5) for finding the fifth term of sequence $\left(a_5\right)$ put n=5 in eq(1)

$a_5=9-5times5$

$⟹a_5=9-25$

$⟹a_5=-16$

.

.

and so on.

hence this sequence form an A.P with first term$a=4$ and commo difference $d=-5$

and the sequence can be written as

$4,-1,-6,-11,-16,....$