Answered By Leonardo F

We have an arithmetic progression, common difference ( $d$d ) 8 and first term ( $t1$t1 ) 5. So:

 $tn=t1+\left(n-1\right)d$tn=t1+(n−1)d 

 $149=5+\left(n-1\right)8$149=5+(n−1)8 

 $149=5+8n-8$149=5+8n−8 

  $149=-3+8n$149=−3+8n

 $152=8n$152=8n 

 $n=\frac{152}{8}=19$n=1528 =19 

So we have 19 terms in the sequence. For the sum of terms: 

 $Sn=\frac{\left(tn+t1\right)n}{2}=\frac{\left(149+5\right)19}{2}$Sn=(tn+t1)n2 =(149+5)192  

 $Sn=1463$Sn=1463 

Answered By Mahboubeh D

The general term is 

t_{n= 4(2n-1)+1}

_{n=1 so} t_{1= 5}

_{n=2 so} t_{2= 13}

_{n=3 so} t_{3= 21}

_{final term is n=19 so} t_{19= 149}

_and s_n=1463

Answered By Kenneth F

The following variables are given:

t1 = 5

d = 8

tn = 149

The general equation of the nth term of an arithmetic progression is:

tn = t1 + (n-1)d 

Replace variables with given values to obtain:

149 = 5 + (n-1)8

Isolate for n by expanding the equation.

149 = 5 + 8n - 8

149 = 8n - 3

152 = 8n

n = 19

149 is the 19th term! 

The equation for the sum of nth variables is Sn = 1/2 (tn + t1)n

Replace variables to obtain:

Sn = 1/2 (149 + 5) x 19 = 1/2 (154) x 19 = 1463