NCERT Exercise 5.1
(1) An Arithmetic Progression (AP) is a list of numbers in which each term is obtained by adding a fixed number ‘d’ to the preceding term, except the first term. The fixed number is called the common difference.
The general term of an AP is
a, a + d, a + 2d, a + 3d, ……….
(2) A given list of numbers a1, a2, a3 ………………… is an AP if the differences a2 - a1, a3 – a2, a4 – a3, ………………… give the same value.
In other words, if a k+1 – ak is the same for different values of k, the term is an AP
(3) In an AP with first term a and common difference d, the nth term is given by
`a_n= a + (n – 1)d`
(4) The sum of the first n terms of an AP:
(5) If l is the last term of the finite AP, say the nth term, then the sum of all terms of the AP is given by:
`S = (n)/(2)(a + l)`
Exercise 5.1 (NCERT)-Part-1
Question (1) In which of the following situations, does the list of numbers involved make an arithmetic progression and why?
(i) The taxi fair after each km when the fare is Rs. 15 for the first km and Rs. 8 for each additional km.
Solution: Fare for 1st km = Rs. 15
Fare for 2nd km `= Rs. 15 + 8 = Rs 23`
Fare for 3rd km `= Rs. 23 + 8 = 31`
Here, each subsequent term is obtained by adding a fixed number (8) to the previous term.
Hence, it is an AP
(ii) The amount of air present in the cylinder when a vacuum pump removes ¼ of the air remaining in the cylinder at a time.
Solution: Let us assume, initial quantity of air = 1
Therefore, quantity removed in first step = ¼
Remaining quantity after first step
Quantity removed in second step
Remaining quantity after second step
Here each subsequent term is not obtained by adding a fixed number to the previous term.
Hence, it is not an AP.
(iii) The cost of dogging a well after every meter of digging, when it costs Rs. 150 for the first meter rises by Rs. 50 for each subsequent meter.
Solution: Cost of digging of 1st meter = 150
Cost of digging of 2nd meter `= 150 + 50 = 200`
Cost of digging of 3rd meter `= 200 + 50 = 250`
Here, each subsequent term is obtained by adding a fixed number (50) to the previous term.
Hence, it is an AP.
(iv) The amount of money in the account every year when Rs. 10000 is deposited at compound interest at 8% per annum.
Solution: Amount in the beginning = Rs. 10000
Interest at the end of 1st year @ 8% `= 10000 xx 8% = 800`
Thus, amount at the end of 1st year `= 10000 + 800 = 10800`
Interest at the end of 2nd year @ 8% `= 10800 xx 8% = 864`
Thus, amount at the end of 2nd year `= 10800 + 864 = 11664`
Since, each subsequent term is not obtained by adding a fixed number to the previous term; hence, it is not an AP.