Respuesta :
Answer:
The sum of first 50 terms is 3425.
Step-by-step explanation:
Let the first tem of A.P be 'a' and it's common difference be 'd'
Thus the 16th term of the A.P is given by
[tex]T_n=a+(n-1)d\\\\40=a+15d..........(i)[/tex]
Now we know that the sum of first 'n' terms of the A.P is given by
[tex]S_n=\frac{n}{2}[2a+(n-1)d]\\\\5=\frac{5}{2}[2a+4d]\\\\1=a+2d..........(ii)[/tex]
Solving equation 'i' and 'ii' simultaneously we get
[tex]40=1-2d+15d\\\\13d=39\\\\\therefore d=3[/tex]
Thus [tex]a=1-2\times 3=-5[/tex]
Thus the sum of first 50 terms euals
[tex]S_{50}=\frac{50}{2}[2\times -5+(50-1)3]\\\\S_{50}=3425.[/tex]
Answer:
[tex]S_{50}=3425[/tex]
Step-by-step explanation:
We have been that the 16th term of an A.P. is 40 and the sum of the first 5 terms is 5.
We will use arithmetic sequence formula and arithmetic sequence sum formula to solve our given problem.
Sequence formula:
[tex]a_n=a_1+(n-1)d[/tex], where,
n = Number of terms in a sequence,
d = Common difference.
[tex]40=a_1+(16-1)d[/tex]
[tex]40=a_1+15d...(1)[/tex]
Sum formula:
[tex]S_n=\frac{n}{2}[2a_1+(n-1)d][/tex]
[tex]5=\frac{5}{2}[2a_1+(5-1)d][/tex]
[tex]5=2.5[2a_1+4d][/tex]
[tex]5=5a_1+10d...(2)[/tex]
Now, we have two unknown and two equations. From equation (1), we will get:
[tex]40-15d=a_1[/tex]
Substitute this value in equation (2).
[tex]5=5(40-15d)+10d[/tex]
[tex]5=200-75d+10d[/tex]
[tex]5=200-65d[/tex]
[tex]200-65d=5[/tex]
[tex]200-200-65d=5-200[/tex]
[tex]-65d=-195[/tex]
[tex]\frac{-65d}{-65}=\frac{-195}{-65}[/tex]
[tex]d=3[/tex]
Substitute [tex]d=13[/tex] in equation (1):
[tex]40=a_1+15(3)[/tex]
[tex]40=a_1+45[/tex]
[tex]40-45=a_1+45-45[/tex]
[tex]-5=a_1[/tex]
Use sum formula to find sum of first 50 terms:
[tex]S_{50}=\frac{n}{2}[2a_1+(n-1)d][/tex]
[tex]S_{50}=\frac{50}{2}[2(-5)+(50-1)3][/tex]
[tex]S_{50}=25[-10+(49)3][/tex]
[tex]S_{50}=25[-10+147][/tex]
[tex]S_{50}=25[137][/tex]
[tex]S_{50}=3425[/tex]
Therefore, the sum of first 50 terms of the given sequence would be 3425.
