A. The sum of the first 10th term is 100.
B. The sum of the nth term is n²
Step-by-step explanation:
Data obtained from the question include:
Sum of 20th term (S20) = 400
Sum of 40th term (S40) = 1600
Sum of 10th term (S10) =..?
Sum of nth term (Sn) =..?
Recall:
Sn = n/2[2a + (n – 1)d]
Sn is the sum of the nth term.
n is the number of term.
a is the first term.
d is the common difference
We'll begin by calculating the first term and the common difference. This is illustrated below:
Sn = n/2 [2a + (n – 1)d]
S20 = 20/2 [2a + (20 – 1)d]
S20= 10 [2a + 19d]
S20 = 20a + 190d
But:
S20 = 400
400 = 20a + 190d (1)
S40 = 40/2 [2a + (40 – 1)d]
S40 = 20 [2a + 39d]
S40 = 40a + 780d
But
S40 = 1600
1600 = 40a + 780d (2)
400 = 20a + 190d (1)
1600 = 40a + 780d (2)
Solve by elimination method
Multiply equation 1 by 40 and multiply equation 2 by 20 as shown below:
40 x equation 1:
40 x (400 = 20a + 190d)
16000 = 800a + 7600. (3)
20 x equation 2:
20 x (1600 = 40a + 780d)
32000 = 800a + 15600d (4)
Subtract equation 3 from equation 4
Equation 4 – Equation 3
32000 = 800a + 15600d
– 16000 = 800a + 7600d
16000 = 8000d
Divide both side by 8000
d = 16000/8000
d = 2
Substituting the value of d into equation 1
400 = 20a + 190d
d = 2
400 = 20a + (190 x 2)
400 = 20a + 380
Collect like terms
400 – 380 = 20a
20 = 20a
Divide both side by 20
a = 20/20
a = 1
Therefore,
First term (a) = 1.
Common difference (d) = 2.
A. Determination of the sum of the 10th term.
First term (a) = 1.
Common difference (d) = 2
Number of term (n) = 10
Sum of 10th term (S10) =..?
Sn = n/2 [2a + (n – 1)d]
S10 = 10/2 [2x1 + (10 – 1)2]
S10 = 5 [2 + 9x2]
S10 = 5 [2 + 18]
S10 = 5 x 20
S10 = 100
Therefore, the sum of the first 10th term is 100.
B. Determination of the sum of the nth term.
First term (a) = 1.
Common difference (d) = 2
Sum of nth term (Sn) =..?
Sn = n/2 [2a + (n – 1)d]
Sn = n/2 [2x1 + (n – 1)2]
Sn = n/2 [2 + 2n – 2]
Sn = n/2 [2 – 2 + 2n ]
Sn = n/2 [ 2n ]
Sn = n²
Therefore, the sum of the nth term is n²
