Find the first term and common difference

Question

The arithmetic sequence a1, a2,...a100 has a sum of 15,000. Find the first term and the common difference if the sum of the terms in the sequence a3, a6, a9... a99 is 5016. Please include solution.

Answer 1

5016 = n(a₃ + a₉₉)/2 using the a-notation from the entire original sequence but only the a3, a6, a9... a99 terms; n=33, so

10,032 = 33(a₃ + a₉₉).  Simplifying, we have

304 =  (a₁+2d + a₁+99d) using the d value from the entire original sequence   ⇐Equation A.

 

15,000 = 100(a₁ + a₁+99d)/2 using the entire sequence, n=100;

simplifying, we get 300 = (a₁ + a₁+99d)  ⇐Equation B.   <.......corrected missing denom.

 

Now you have two equations, two unknowns, can solve for a₁ and d.  YOU FINISH! AND CHECK!!

 

Answer 2

 

a1+a2+a3+...+a98+a99+a100=15,000

a1+a100=a2+a99=a3+a98=.....

there are 50 of these sums and they are all equal

15,000/50=300

there are 50 sums of 300

 

a3+a6+a9a+...+a93+a96+a99=5016

there are 33 terms

there is an odd number of terms so there will be 16 sums that are equal and a middle term that is 1/2 any of the sums

in other words, there are 16 1/2 sums

5016/16.5=304

there are 16 sums of 304 and a middle term of 152

 

a3+a99=304

a6+a96=304

and so on

 

from the top...

a3+a98=300 along with what we just found...

a3+a99=304

subtract the top equation from the bottom equation to get...

a99-a98=4

the common difference is d=4

now find the first term...

15,000=(100/2)(2a+[99*4])

15,000=50(2a+396)

15,000=100a+19,800

15,000-19,800=100a

-4800=100a

a=-4800/100

a=-48, the first term

 

the sequence is -48, -44, -40,...,340, 344, 348

 

 

 

Answer 3

Arithmetic sequence:

{a₁, a₂, a₃, a₄, ... a_n} corresponds to:

{a+(0)d, a+(1)d, a+(2)d, a+(3)d, ... a+(n-1)d}

 

Where;

d = common difference

n = the number of the term

 

The equation for the sum of the first n-terms of an arithmetic sequence is:

 

S_n = n(a₁ + a_n)/2

 

Given the sum of every 3rd term in the sequence: a₃, a₆, a₉... a₉₉ is 5016

 

(a+2d) + (a+5d) + (a+8d) + .... (a+98d) = 5016

 

We can see that adding up every 3rd "a" to the 99th term will give 99/3 = 33a

 

The coefficients of "d" make their own arithmetic sequence of 33 numbers: a = 2 and d = +3

{2, 5, 8, ...98} = {2+(0)3, 2+(1)3, 2+(2)3, .. }

The sum of coefficients of d then = 33[2 + 2+(32)3]/2

33(100)/2 = 1650

 

So our Original sum of every 3rd tem:

5016 = 33a + 1650d

 

We are also given the sum of the first 100 terms equals 15,000

 

a₁ = a + (0)d = a 

a₁₀₀ =  a + (100-1)d = a + 99d

 

15,000 = 100[(a)+ (a + 99d)] / 2

15,000 = 50(2a + 99d)

300 = 2a + 99d

 

We need to use substitution to solve for the common difference, d in the two equations we've come up with because we have two unknown variables (a and d).

 

5016 = 33a + 1650d

300 = 2a + 99d

 

Rearranging the second Equation to equal a.

 

300 - 99d = 2a

(300 - 99d)/2 = a

 

substitute into the 1st Equation.

 

5016 = 33(300-99d)/2 + 1650d

5016 = 4950 - 1633.5d  + 1650d

combine like terms

66 = 16.5d

4 = d