Question 1004534



you are given that sum of 4 terms is 38 and sum of 8 terms is 124.


the formula for sum of an arithmetic sequence is Sn = n/2 * (A1 + An)


the formula for the nth term of an arithmetic sequence is An = A1 + (n-1)d


A1 is the first term.
An is the nth term.
n is the number of terms.
d is the common difference.


since An is equal to A1 + (n-1)d, you can replace An in the formula of Sn = n/2 * (A1 + An with A1 + (n-1)d to get:


Sn = n/2 * (A1 + An) becomes:


Sn = n/2 * (A1 + A1 + (n-1)d) which then becomes:


Sn = n/2 * (2A1 + (n-1)d)


it is this last formula that will help you find what you are looking for.


you are given that the sum of the first 4 terms in the sequence is equal to 34.


the summation formula for that becomes:


38 = 4/2 * (2A1 + 3d)


you can simplify this formula to get:


38 = 4A1 + 6d


you are also given tha the sum of the first 8 terms in the sequence is equal to 124.


the summation formula for that becomes:


124 = 8/2 * (2A1 + 7d)


You can simplify this to get:


124 = 8A1 + 28d


you now have 2 formulas that you need to solve simultaneously.


they are:


38 = 4A1 + 6d
124 = 8A1 + 28d


multiply both sides of the first equation by 2 and bring down the second equation as is to get:


76 = 8A1 + 12d
124 = 8A1 + 28d


subtract the first equation from the second equation to get:


48 = 16d


solve for d to get:


d = 3


now that you know d = 3, you can use either original equation to solve for A1.


your original equations to use are:


38 = 4A1 + 6d
124 = 8A1 + 28d


you will get A1 = 5.


your solution is that A1 = 5 and d = 3


A1 is the first term.
d is the common difference.