Answer:
The first term is 5
The common difference is 3
Explanation:
Let x be the first term. Let y be the common difference between each number in the sequence. x and the next three terms would be:
x, x+y, x+2y, and x+3y
The sum of the 4 terms is 4x + 6y and is equal to 38
4x + 6y = 38
4x = 38 - 6y
x = (19/2) - (3/2)y [x is isolated here, to the left, for use in a lovely substitution coming up]
or x = 9.5 - 1.5y [simplified]
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The sum of the first 7 terms would be the first 4 [from above: 4x + 6y] plus the next 3 terms;
4x + 6y
x + 4y
x + 5y
x + 6y
7x + 21y
7x + 21y is equal to 98
7x + 21y = 98
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We have two equations and two unknowns, so we should be able to find an answer by substitution:
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From above:
x = (19/2) - (3/2)y
7x + 21y = 98
Now use the first definition of x in the second equation:
7x + 21y = 98
7( (19/2) - (3/2)y) + 21a = 98
66.5 - 10.5y + 21y = 98
10.5y = 31.5
y = 3
Now use this value of y in either equation to find x:
7x + 21*(3) = 98
7x + 63 = 98
7x = 35
x = 5
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x is the first term: 5
y is the common difference: 3
Check:
Do the first 4 terms sum to 38?
5 + 8 + 11 + 14 = 38 YES
Do the first 7 terms sum to 98?
38 + 17 + 20 + 23 = YES