help needed in solving maths problem_arithmetic progression?

Question

solve the equation 3 + 7 + 11 + 15 ........ + X = 1830

tanx
tom

Answer 1

I have no clue but thought I would give you a heads up on a very good web site.
www.math.com

Answer 2

Given, 3+7+11+15....+X=1830
Then, First term (T1) = 3 ; Last term (Tn) = X ; and Sum of the given series (Sn) = 1830, also it is known by simple inspection of the series that d = 4
Here suitable equation to find `n' the number of terms is
Sn = n/2 [2T1 + (n-1)d]
Substituting the known values we have,
1830 = n[2n +1]
Prime factors of 1830 are 2*5*3*61
Then, 2*5*3*61 = n(2n + 1)
Transforming the LHS so as to resemble the RHS, we have,
30(2*30 +1) = n (2n + 1)
On comparison of LHS and RHS, we get
n=30
Now it is easy to find the last term using the equation
Tn=X = T1+(n-1)d
ie X = 3 + 29*4 =119

Answer 3

Let there be n numbers
nth number = 4n -1
sum of them 4n(n+1)/2 -n
= 2n(n+1) -n = 2n**2 +n = 1830
solving this n = 30 so last number is the 30th term x = 4*30-1 = 119
I hope it clarifies

Answer 4

sum=n/2[2a+(n-1)d]
substituting n/2[6+(n-1)4]=1830
n[6+4n-4]=3660
4n^2+2n-3660=0
=>2n^2+n-1830=0 factorising n=30
n th term=a+(n-1)d
substituting X=3+(30-1)4=3+116=119

Answer 5

It took me a lengthy time period to parent this out. a million=4+4/4+4 2=4/4+4/4 3=(4+4+4)/4 4=(4-4)4+4 5=(4x4+4)/4 6=(4+4)/4+4 7=sq. root of four +4/4 +4 8=(4x4)-(4+4) 9=4/4+4+4 10=sq. root of four +sq. root of four +sq. root of four +4 11=(4!+4!-4)/4 12=sq. root of four +sq. root of four +4+4 13=(4!+4!+4)/4 14=sq. root of four x4 +4 +sq. root of four 15=4x4-(4/4) 16=(4-4)+(4x4) 17=4x4+4/4 18=4x4-sq. root of four +4 19=4! -4 -4/4 20=(sq. root of four +4)4-4

Answer 6

maybe without the dots it would be easier.............