Given an arithmetic progression whose 10th term is 10 amd the sum of the first 10 terms is -35, determine the first term and the common difference
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Given an arithmetic progression whose 10th term is 10 amd the sum of the first 10 terms is -35, determine the first term and the common difference
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GIVEN:
a10 = 10
S10 = -35
SOLUTION:
L = a1 + (n-1)(d)
where:
L = last term or nth term
a1 = first term
n = number of terms
d = common difference
---------
10 = a1 + (10-1)(d)
10 = a1 + 9d
a1 = 10 - 9d -------> eq.1
S = (n/2)[2(a1) + (n-1)(d)]
where:
S = sum of all terms
n = number of terms
a1 = first term
d = common difference
---------
-35 = (10/2)[2(a1) + (10-1)(d)]
-35 = (5)[2(a1) + 9d]
-35 = 10(a1) + 45d -------> eq. 2
Substitute eq. 1 to eq. 2:
-35 = 10(10 - 9d) + 45d
-35 = 100 - 90d + 45d
-35 = 100 - 45d
45d = 100 + 35
45d = 135
divide both sides by 45:
d = 3
therefore, the common difference is 3.
Substitute the value of d to eq. 1:
a1 = 10 - 9d
a1 = 10 - 9(3)
a1 = 10 - 27
a1 = -17
the value of first term is -17.