Question

common difference of an arithmetic sequence is 5 and its 21st term 108. a)what is its 22th term?
b)what is the sum of first 43 terms of this sequence?​

Answer 1

Given:

The common difference of an arithmetic sequence is 5 and its 21st term is 108.

To Find:

a) What is its 22nd term?

b) What is the sum of its first 43 terms?

Solution:

First, we know that the nth term of an AP can be expressed as

                                   $T_{n} =a+(n-1)d$

and the common difference is given $d$=5

now it is given that the 21st term is 108, putting this on the formula we get,

                                $108=a+(21-1)5\\108=a+20*5\\a=108-100\\a=8$

so we get first term $a$=8

now we have the following information with us,

$a$=8

$d$=5

moving further

a) 22nd term

using the formula

$T_{22} =a+(22-1)d\\=8+21*5\\=8+105\\=113$

Hence, the 22nd term is 113.

b) sum of first 43 terms

For a summation of the arithmetic sequence we use the formula as,

                              $S_{n} =\frac{n}{2} (2a+(n-1)d)$

now finding the sum of the first 43 terms,

                            $S_{43}=\frac{43}{2} (2*8+(43-1)5)\\=\frac{43}{2} (16+210)\\=\frac{43}{2}*226\\=4859$

Hence, the sum of the first 43 terms will be 4859.