Answer:
a) [tex] t_n = (log x + log √2)n[/tex]
b) [tex] S_{40} =20\bigg(log x + log \sqrt 2\bigg)[ 1+n][/tex]
Step-by-step explanation:
Given arithmetic sequence is:
log x + log √2, 2 log x + log 2, 3 log x + log 2√2......
First term a = log x + log √2
common difference d = 2 log x + log 2 - (log x + log √2)
= 2 log x + log 2 - log x - log √2
= log x + log 2 - log √2
= log x + log (√2*√2)- log √2
= log x + log √2 + log √2- log √2
= log x + log √2
(a)
General term of the arithmetic sequence is given as:
[tex]t_n = a + (n - 1)d[/tex]
[tex]\implies t_n = (log x + log \sqrt 2) + (n - 1)( log x + log √2 )[/tex]
[tex]\implies t_n = (log x + log \sqrt 2)[1+ + (n - 1)][/tex]
[tex]\implies t_n = (log x + log \sqrt 2)[1+ n - 1][/tex]
[tex]\implies t_n = (log x + log \sqrt 2)[n][/tex]
[tex]\implies \red{\bold{t_n = (log x + log \sqrt 2)n}}[/tex]
(b)
Sum of first n terms of an arithmetic sequence is given as:
[tex]S_n =\frac{n}{2}\bigg(a+t_n\bigg)[/tex]
[tex]\implies S_{40} =\frac{40}{2}\bigg[ (log x + log \sqrt 2)+(log x + log \sqrt 2)n\bigg][/tex]
[tex]\implies S_{40} =20\bigg(log x + log \sqrt 2\bigg)(1+n)[/tex]
