What is the common difference of this arithmetic sequence?

Question

In an arithmetic sequence with 94 terms, the 8th term is 63, the sum of terms is 1000. What is the common difference?

 

Answer 1

An arithmetic sequence is of the form ai+b where i=0,1,2,3......n-1

 

In the question, the number of terms, n=94

 

The common difference which is unknown is the coefficient, a.  Why? Consider the

difference between any consecutive terms, let's say the j+1 th and the j th.

 

[a(j)+b]  -  [a(j-1)+b]  =a

 

We are also told the 8th term is  63 so:  7a+b=63     equation 1

 

And the sum of terms is 1000 so:    ∑ ai+b  (i=0 to 93)=1000

 

Which can be simplified as follows:  ∑ ai+∑b=a∑i+∑b=a∑i+94b=1000

 

And ∑i (i=0 to n-1)=n(n-1)/2  a formula you should memorize.  (Gauss is a famous practitioner aged 9: http://www.wyzant.com/resources/blogs/277085/mathematical_journeys_carl_gauss_and_the_sum_of_an_arithmetic_series

 

So  a*94*93/2+94b=1000 or 94(b+a*93/2)=1000  equation 2


All we need to do now is solve the two simultaneous linear equations.  We are looking for a, so it will be quicker if we substitute for b.  From equation 1, b=(63-7a)

 

So  94(63+79a/2)=1000

 

a=(1000/94-63)*2/79=(1000-63*94)/(79*47)

 

a=-4992/3713

Answer 2

Hi Shinee,

 

Let a be the 1st term,d is the commom difference,

nth term is a+(n-1)d

s_n = n/2 {2a+(n-1)d}

 

n=94, t₈ =63, s₉₄ = 1000.

 

t₈ =a+7d = 63......1

s₉₄ =94/2{2a+93d} = 47(2a+93d)=1000.......2

from 1, we get a=63-7d, substituting in 2,

 

47{2(63-7d)+93d} =1000

47( 126 -14d +93d) =1000

47(126+79d) =1000

5922 +3713d=1000

3713d= -4922

d= -4922/3713.

Pl check question.