Question

Given the recursive formula for an arithmetic sequence find the common difference and the 52nd term

Answer 1

Answer:

              Common difference(d)                  $a_{52}$

(21)                   -10                                          -548

(22)                   -7                                          -323

(23)                    10                                           547

(24)                   -100                                       -5118

Step-by-step explanation:

Let the common difference be denoted by 'd'.

Also the nth difference of an arithmetic sequence is given by: $a_{n}=a_{1}+(n-1)\times d$

(21)

We are given a recursive formula as:

$a_{n}=a_{n-1}-10$

The first term is given by:

$a_{1}=-38$

The common difference for an arithmetic sequence is given by:

$a_{n}-a_{n-1}$

Hence, here we have the common difference as:

$d=-10$

The nth term of an arithmetic sequence is given by:

$a_{n}=a_{1}+(n-1)\times d$

Here $a_{1}=-38$ and $d=-10$.

Hence, $a_{52}=-38+(52-1)\times (-10)$

Hence, $a_{52}=-548$

(22)

$a_{n}=a_{n-1}-7$

$a_{1}=34$

The common difference for an arithmetic sequence is given by:

$a_{n}-a_{n-1}$

Hence, here we have the common difference as:

$d=-7$

Here $a_{1}=34$ and $d=-7$.

Hence, $a_{52}=34+(52-1)\times (-7)$

Hence, $a_{52}=-323$

(23)

$a_{n}=a_{n-1}+10$

$a_{1}=37$

The common difference for an arithmetic sequence is given by:

$a_{n}-a_{n-1}$

Hence, here we have the common difference as:

$d=10$

Here $a_{1}=37$ and $d=10$.

Hence, $a_{52}=37+(52-1)\times (10)$

Hence, $a_{52}=547$

(24)

$a_{n}=a_{n-1}-100$

$a_{1}=-18$

The common difference for an arithmetic sequence is given by:

$a_{n}-a_{n-1}$

Hence, here we have the common difference as:

$d=-100$

Here $a_{1}=-18$ and $d=-100$.

Hence, $a_{52}=-18+(52-1)\times (-100)$

Hence, $a_{52}=-5118$