![Steps Solution
2. Find the amount received by each
person using the formula Sn =
𝑛
2
[2A1 +
(n – 1)d].
Sn =
𝑛
2
[2A1 + (n – 1)d]
Sn =
6
2
[ 2A1 + (6 – 1)(-50,000 ]
2,000,000 = 3 [ 2A1 + (5)(-50,000) ]
2,000,000 = 3[2A1 -250,000]
2,000,000 = 6A1 – 750,000
2,000,000 + 750,000 = 6A1
2,750,000 = 6A1
458,333.33 = A1](https://image.slidesharecdn.com/problemsolvinginvolvingarithmeticsequencesandseries-201017133232/85/Problem-solving-involving-arithmetic-sequences-and-series-4-320.jpg)
![Steps Solution
3. The amount of savings in a year can
be computed using the formula
Sn =
𝑛
2
[ 2A1 + (n – 1)d ]
Sn =
𝑛
2
[ 2A1 + (n – 1)d ]
n = 52; d = 10; A1 = 150
S52 =
52
2
[ 2(150) + (52 – 1)10 ]
= 26[300 + 51(10)]
= 26 [300 + 510]
= 26(810)
= 21, 060
So, the total amount of savings in one year is Php 21, 060.](https://image.slidesharecdn.com/problemsolvinginvolvingarithmeticsequencesandseries-201017133232/85/Problem-solving-involving-arithmetic-sequences-and-series-7-320.jpg)
Problem solving involving arithmetic sequences and series
- 2. Arithmetic sequences and series are useful in solving some real-life problems.
- 3. Example 1. The amount of Php 2,000,000 is distributed among six people so that each person after the first receives Php 50,000 less than the preceding person. How much does each person receive? Steps Solution 1. Identify the given information Let A1, A2, A3,…..A6 be the amount to be distributed among six people. So, A1 + A2 + A3 + A4 + A5 + A6 = S6 Common difference is -50,000 Let A1 be the amount received by the first person
- 4. Steps Solution 2. Find the amount received by each person using the formula Sn = 𝑛 2 [2A1 + (n – 1)d]. Sn = 𝑛 2 [2A1 + (n – 1)d] Sn = 6 2 [ 2A1 + (6 – 1)(-50,000 ] 2,000,000 = 3 [ 2A1 + (5)(-50,000) ] 2,000,000 = 3[2A1 -250,000] 2,000,000 = 6A1 – 750,000 2,000,000 + 750,000 = 6A1 2,750,000 = 6A1 458,333.33 = A1
- 5. Steps Solution 3. Since each person after the first receives Php 50,000 less than the preceding person and A1 = 475,000, then we can already determine the amount received by each one of them. A1 = 458,333.33 A2 = A1 – 50,000 =408,333.33 A3 = A2 – 50,000 = 358,333.33 A4 = A3 – 50,000 = 308,333.33 A5 = A4 – 50,000 = 258,333.33 A6 = A5 – 50,000 = 208,333.33 Therefore, the amount received by each person in sequence form is as follows: Php458,333.33, Php408,333.33, Php358,333.33, Php308,333.33, Php258,333.33 and Php208,333.33.
- 6. Example 2. Suppose you have Php 150 for one week and that each week thereafter you save Php 10 more than the preceding week. How much will you have saved by the end of one year? Steps Solution 1. Identify the given information Let A1 = Php150 be the initial amount saved in one week Common difference = 10 2. There are 52 weeks in a year, so the nth term of the series is A52 150 + 160 + … + A52
- 7. Steps Solution 3. The amount of savings in a year can be computed using the formula Sn = 𝑛 2 [ 2A1 + (n – 1)d ] Sn = 𝑛 2 [ 2A1 + (n – 1)d ] n = 52; d = 10; A1 = 150 S52 = 52 2 [ 2(150) + (52 – 1)10 ] = 26[300 + 51(10)] = 26 [300 + 510] = 26(810) = 21, 060 So, the total amount of savings in one year is Php 21, 060.
- 8. THANK YOU!