# Arithmetic Progressions - Class 10 CBSE Math Solutions of Chapter 5, NCERT Mathematics Textbook Exercise 5.1

## Chapter 5, ARITHMETIC PROGRESSIONS

### NCERT Solutions for Mathematics Textbook Exercise 5.1

(Page 99)
1: In which of the following situations, does the list of numbers involved make an arithmetic progression (AP), and why?
i.    The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.
ii.    The amount of air present in a cylinder when a vacuum pump removes ¼ of the air remaining in the cylinder at a time.
iii.    The cost of digging a well after every meter of digging, when it costs Rs 150 for the first meter and it rises by Rs 50 for each subsequent meter.
iv.    The amount of money in the account every year, when Rs 10000 is deposited at compound interest @ 8% per annum.
Solution.1:
(i) At the end of first km the fare is Rs 15 and going further with addition of each km the fares will be Rs 23, Rs 31, Rs 39 … obtained by adding Rs 8 to its preceding term.
So, the answer is Yes since 15, 23, 31, 39, … forms an AP where a = 15 and d = 8.

(ii) Let V is the initial amount of air present in cylinder. So, the amount of air that will remain in the cylinder after each cycle of removing air by vacuum pump will be V, ¾V , (¾)2V, …
The answer is No since the above sequence does not form an AP.

(iv) [Remember with ‘compound interest’ the money growing (left after every year) in an account can not form AP]
The answer is No since amount left after every year will be -
10000 (1 + 8/100), 10000 (1 + 8/100)2, 10000 (1 + 8/100)3, …. which is not an Arithmetic Progression.

2: Write first four terms of AP (Arithmetic Progression), when the ‘first term’ a and the ‘common difference’ d are given as follows:
(i) a = 10, d = 10                (ii) a = –2, d = 0          (iii) a = 4, d = –3
(iv) a = –1, d = ½               (v) a = –1.25, d = –0.25
Solution.2:
(i) The AP is -
a, a+d, a+2d, a+3d
or, 10, 10+10, 10+2(10), 10+3(10)
The answer is 10, 20, 30, 40.

(ii) The AP is -
a, a+d, a+2d, a+3d
or, –2, –2+0, –2+2(0), –2+3(0)
The answer is –2, –2, –2, –2.

(iii) Try it yourself.

(iv) The AP is expressed by -
a, a+d, a+2d, a+3d
or, –1, –1+1/2, –1+2 x 1/2, –1+3 x ½
or, –1, –1/2, –1+1, –1+3/2
The answer is –1, –1/2, 0, 1/2.

(v) Try it yourself.

3: For the following Arithmetic Progressions, write the first term and the common difference:
(i) 3, 1, –1, –3, …              (ii) –5, –1, 3, 7, …
(iii) 1/3, 5/3, 9/3, 13/3, …   (iv) 0.6,  1.7, 2.8, 3.9, …
Solution.3:
(i) We have,
a1 = 3, a2 = 1, a3 = –1, a4 = –3.
So we get common difference by
a2 - a1 = 3 - 1 = –2
a3 - a2 = –1 - 1 = –2
a4 - a3 = –3 - (–1) = –2
Here, first term (a) = 3 and common difference (d) = –2

(ii) We have,
a1 = –5, a2 = –1, a3 = 3, a4 = 7.
So we get common difference by
a2 - a1 = –1 - (–5) = 4
a3 - a2 = 3 - (–1) = 4
a4 - a3 = 7 - 3 = 4
Here, first term (a) = –5 and common difference (d) = 4

(iii) (iv) Taking hint from the above try and do it yourself.

(Page 100)
4: Which of the following are APs? If they form an AP, find the common difference d and write three common terms.
(i) 2, 4, 8, 16, …          (ii) 2, 5/2, 3, 7/2, …    (iii) –1.2, –3.2, –5.2, –7.2, …
(iv) –10, –6, –2.2, …  (v) 3, 3+√2, 3+2√2, 3+3√2, … (vi) 0.2, 0.22, 0.222, 0.2222, …
(vii) 0, –4, –8, –12, …    (viii) –1/2, –1/2, –1/2, –1/2, …(ix) 1, 3, 9, 27, …
(x) a, 2a, 3a, 4a, … (xi) a, a2, a3, a4, …   (xii) √2, √8, √18, √32, …
(xiii) √3, √6, √9, √12, … (xiv) 12, 32, 52, 72, …  (xv) 12, 52, 72, 73, …
Solution.4:
(i) We have,
a1 = 2, a2 = 4, a3 = 8, a4 = 16.
So we get common difference by subtracting
a2 - a1 = 4 - 2 = 2
a3 - a2 = 8 - 4 = 4
a4 - a3 = 16 - 8 = 8
Since the difference is not constant so, they do not form AP.

(ii) We have,
a1 = 2, a2 = 5/2, a3 = 3, a4 = 7/2.
So we get common difference by subtracting
a2 - a1 = 5/2 - 2 = 1/2
a3 - a2 = 3 - 5/2 = 1/2
a4 - a3 = 7/2 - 3 = 1/2
So, the given list of numbers are in arithmetic progression with common difference (d) = ½. The next three terms are:
7/2 + 1/2 = 4
4 + 1/2 = 9/2
9/2 + 1/2 = 5
The answer is 4, 9/2 and 5.

(iii) & (iv) Do it yourself.

(v) We have,
a1 = 3, a2 = 3+√2, a3 = 3+2√2, a4 = 3+3√2.
So we get common difference by
a2 - a1 = 3+√2 - 3 = √2
a3 - a2 = 3+2√2 - 3+√2 = √2
a4 - a3 = 3+3√2 - 3+2√2 = √2
So, the given list of numbers are in AP with common difference (d) = √2. The next three terms are:
3+3√2 + √2 = 3+4√2
3+4√2 + √2 = 3+5√2
3+5√2 + √2 = 3+6√2
The answer is 3+4√2, 3+5√2 and 3+6√2.

(vi), (vii), (viii) Following the above process try yourself.

(ix) We have,
a1 = 1, a2 = 3, a3 = 9, a4 = 27.
Now,
a2 - a1 = 3 - 1 = 2
a3 - a2 = 9 - 3 = 6
a4 - a3 = 27 - 9 = 18
Since the difference is not constant so, they are not in AP.

(x) Try to do it yourself.

(xi) We have,
a1 = a, a2 = a2, a3 = a3, a4 = a4.
Now by subtracting
a2 - a1 = a2 - a = a(a -1)
a3 - a2 = a3 - a2 = a2(a -1)
a4 - a3 = a4 - a3 = a3(a -1)
Since the difference is not constant so, they do not form AP.

(xii) We have,
a1 = √2, a2 = √8, a3 = √18, a4 = √32.
So we get common difference by
a2 - a1 = √8 - √2 = 2√2 - √2 = √2
a3 - a2 = √18 - √8 = 3√2 - 2√2 = √2
a4 - a3 = √32 - √18 = 4√2 - 3√2 = √2
So, the given list of numbers are in AP with common difference (d) = √2. The next three terms are:
√32 + √2 = 4√2 + √2 = 5√2 = √50
√50 + √2 = 5√2 + √2 = 6√2 = √72
√72 + √2 = 6√2 + √2 = 7√2 = √98
The answer is √50, √72 and √98.

Taking hints from these solutions try to solve remaining problems. At the end of completing all Exercises from each Chapter given in NCERT Math textbook, we shall provide CBSE Notes (Hints) / CBSE Solutions, Mathematics Sample Questions with their Solutions and many more…
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