Question

If the sum of the first n terms of the series 4+7+10+..... is 209, find n​

Answer 1

Answer:

Step-by-step explanation:

wow, this is kinda complicated math for HS..  :?   we have to know what type of series this is, and that means knowing all the different kinds.. unless maybe in the instructions it tells you you are working with arithmetic series? It is one of those so we can use those formulas... other series can have much more complex formulas..  series comes up in Calc II in college.. but .. unless you're going to be an engineer you probably don't need it..  soooooo anyway..

we can use  Sn = n/2(2a1 + d(n-1) )

where d is the difference of a2 -a1 ( this will be the same for all terms)

since we're given Sn = 209 set the problem to 209 and solve for n  :/

sounds hard huh.. :?  let's see

209 = n/2( 2(4) + 3(n-1))

209 = n/2(8 + 3n-3)

209 = n/2(5+3n)

2*209 = n(5+3n)

418 = 3$n^{2}$ + 5n

0 = 3$n^{2}$ + 5n-418    ( use quadratic formula to find n )

( -5 ± $\sqrt{5^{2}-4*3*(-418) }$  ) / 2(3)

( -5 ±$\sqrt{25+5016}$ ) / 6

( -5 ±$\sqrt{5041}$ ) / 6

( -5 ± 71 ) / 6  ( they were nice to us and make it work out perfectly :)  )

positive route =  71-5 /6 = 66/6 =11

negative route = 71-(-5) /6  = 71+5/6 = 76/6 = 12$\frac{2}{3}$

since 12$\frac{2}{3}$  isn't going to really work for us... let's try 11

an= a1 + d(n-1)

an = 4 + 3(11-1)

an = 4 + 30

an = 34

now let's see if we can get 209 for our 11 terms of the series

Sn = 11/2(2(4) + 3(11-1))

Sn = 11/2(8 + 30)

Sn = 418 /2

Sn = 209  yay.. we got it.. it's the 11th term.. and thank you  quadratic formula.  You get a Christmas present this year  :P

n = 11

Answer 2

Answer:

n=11

Step-by-step explanation:

n/2 (2*4 + (n-1)*3) = 209

1/2 (3n^2+5n) = 209

3n^2+5n-418 = 0

(3n+38)(n-11) = 0

n=11

Hope it helps you .....