Mathematics · Arithmetic Progressions · NCERT Exercises
Exercise 5.1
Complete, independently verified solutions for NCERT Exercise 5.1.
Change resourceExercise 5.1
Mathematics · Chapter 5 · NCERT Exercises
Question
Question 1
Decide whether these situations form APs and why: (i) taxi fare ₹15 first km plus ₹8 each extra km; (ii) a pump removes 1/4 of remaining air each time; (iii) well digging costs ₹150 first metre plus ₹50 each next metre; (iv) ₹10000 at 8% compound interest.
Solution
Board-exam working:
(i) Taxi fare: Fare for 1 km=₹15.
Fare for 2 km=₹15+₹8=₹23.
Fare for 3 km=₹23+₹8=₹31.
Therefore, the list of fares is 15,23,31,….
Difference between the second and first terms=23−15=8.
Difference between the third and second terms=31−23=8.
Since the consecutive differences are equal, the taxi fares form an AP with common difference d=8.
(ii) Air remaining in the cylinder: Let the initial quantity of air be V.
After the first operation, air remaining=V−V/4=3V/4.
After the second operation, air remaining=3V/4−(1/4)(3V/4)=3V/4−3V/16=9V/16.
After the third operation, air remaining=9V/16−(1/4)(9V/16)=9V/16−9V/64=27V/64.
Therefore, the list is V,3V/4,9V/16,27V/64,….
Second term−first term=3V/4−V=−V/4.
Third term−second term=9V/16−3V/4=9V/16−12V/16=−3V/16.
Since −V/4≠−3V/16, the consecutive differences are not equal.
Hence the list is not an AP.
(iii) Cost of digging the well: Cost of the first metre=₹150.
Cost of the second metre=₹150+₹50=₹200.
Cost of the third metre=₹200+₹50=₹250.
Therefore, the list of costs is 150,200,250,….
Difference between the second and first terms=200−150=50.
Difference between the third and second terms=250−200=50.
Since the consecutive differences are equal, the costs form an AP with common difference d=50.
(iv) Compound interest: Principal=₹10000 and rate=8% per annum.
Amount after the first year=10000+(8/100)×10000=10000+800=₹10800.
Amount after the second year=10800+(8/100)×10800=10800+864=₹11664.
Amount after the third year=11664+(8/100)×11664=11664+933.12=₹12597.12.
Therefore, the list of yearly amounts is 10000,10800,11664,12597.12,….
Second term−first term=10800−10000=800.
Third term−second term=11664−10800=864.
Since 800≠864, the consecutive differences are not equal.
Hence the list is not an AP.
Final answer
(i) AP, d=8; (ii) not an AP; (iii) AP, d=50; (iv) not an AP.
Question
Question 2
Write four terms for: (i) a=10,d=10; (ii) a=−2,d=0; (iii) a=4,d=−3; (iv) a=−1,d=1/2; (v) a=−1.25,d=−0.25.
Solution
Board-exam working:
To obtain each next term, add the common difference d to the preceding term.
(i) Given a=10 and d=10.
First term: a₁=10.
Second term: a₂=a₁+d=10+10=20.
Third term: a₃=a₂+d=20+10=30.
Fourth term: a₄=a₃+d=30+10=40.
Hence the first four terms are 10,20,30,40.
(ii) Given a=−2 and d=0.
First term: a₁=−2.
Second term: a₂=−2+0=−2.
Third term: a₃=−2+0=−2.
Fourth term: a₄=−2+0=−2.
Hence the first four terms are −2,−2,−2,−2.
(iii) Given a=4 and d=−3.
First term: a₁=4.
Second term: a₂=4+(−3)=1.
Third term: a₃=1+(−3)=−2.
Fourth term: a₄=−2+(−3)=−5.
Hence the first four terms are 4,1,−2,−5.
(iv) Given a=−1 and d=1/2.
First term: a₁=−1.
Second term: a₂=−1+1/2=−1/2.
Third term: a₃=−1/2+1/2=0.
Fourth term: a₄=0+1/2=1/2.
Hence the first four terms are −1,−1/2,0,1/2.
(v) Given a=−1.25 and d=−0.25.
First term: a₁=−1.25.
Second term: a₂=−1.25+(−0.25)=−1.50.
Third term: a₃=−1.50+(−0.25)=−1.75.
Fourth term: a₄=−1.75+(−0.25)=−2.00.
Hence the first four terms are −1.25,−1.50,−1.75,−2.00.
Final answer
(i) 10,20,30,40; (ii) −2,−2,−2,−2; (iii) 4,1,−2,−5; (iv) −1,−1/2,0,1/2; (v) −1.25,−1.50,−1.75,−2.00.
Question
Question 3
For each AP state a and d: (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
Board-exam working:
For an AP, the first term is a and the common difference is d=a₂−a₁.
We verify d using every consecutive pair shown.
(i) The AP is 3,1,−1,−3,….
Therefore a=3.
a₂−a₁=1−3=−2.
a₃−a₂=−1−1=−2.
a₄−a₃=−3−(−1)=−3+1=−2.
All differences are equal.
Hence d=−2.
(ii) The AP is −5,−1,3,7,….
Therefore a=−5.
a₂−a₁=−1−(−5)=−1+5=4.
a₃−a₂=3−(−1)=3+1=4.
a₄−a₃=7−3=4.
All differences are equal.
Hence d=4.
(iii) The AP is 1/3,5/3,9/3,13/3,….
Therefore a=1/3.
a₂−a₁=5/3−1/3=(5−1)/3=4/3.
a₃−a₂=9/3−5/3=(9−5)/3=4/3.
a₄−a₃=13/3−9/3=(13−9)/3=4/3.
All differences are equal.
Hence d=4/3.
(iv) The AP is 0.6,1.7,2.8,3.9,….
Therefore a=0.6.
a₂−a₁=1.7−0.6=1.1.
a₃−a₂=2.8−1.7=1.1.
a₄−a₃=3.9−2.8=1.1.
All differences are equal.
Hence d=1.1.
Final answer
(i) a=3,d=−2; (ii) a=−5,d=4; (iii) a=1/3,d=4/3; (iv) a=0.6,d=1.1.
Question
Question 4
Determine which form APs, give d and three next 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 repeated (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
Board-exam working:
A list is an AP only when all its consecutive differences are equal.
(i) 4−2=2, 8−4=4 and 16−8=8.
Since 2,4 and 8 are unequal, the list is not an AP.
(ii) 5/2−2=5/2−4/2=1/2.
3−5/2=6/2−5/2=1/2.
7/2−3=7/2−6/2=1/2.
The differences are equal, so it is an AP with d=1/2.
The next terms are 7/2+1/2=4, 4+1/2=9/2 and 9/2+1/2=5.
(iii) −3.2−(−1.2)=−3.2+1.2=−2.
−5.2−(−3.2)=−5.2+3.2=−2.
−7.2−(−5.2)=−7.2+5.2=−2.
The differences are equal, so it is an AP with d=−2.
The next terms are −7.2−2=−9.2, −9.2−2=−11.2 and −11.2−2=−13.2.
(iv) −6−(−10)=4, −2−(−6)=4 and 2−(−2)=4.
Therefore it is an AP with d=4.
The next terms are 2+4=6, 6+4=10 and 10+4=14.
(v) (3+√2)−3=√2.
(3+2√2)−(3+√2)=√2.
(3+3√2)−(3+2√2)=√2.
Therefore it is an AP with d=√2.
The next terms are 3+4√2, 3+5√2 and 3+6√2.
(vi) 0.22−0.2=0.02, 0.222−0.22=0.002 and 0.2222−0.222=0.0002.
Since the differences are unequal, the list is not an AP.
(vii) −4−0=−4, −8−(−4)=−4 and −12−(−8)=−4.
Therefore it is an AP with d=−4.
The next terms are −12−4=−16, −16−4=−20 and −20−4=−24.
(viii) Every term is −1/2.
Thus (−1/2)−(−1/2)=0 for every consecutive pair.
Therefore it is an AP with d=0.
The next three terms are −1/2,−1/2,−1/2.
(ix) 3−1=2, 9−3=6 and 27−9=18.
Since the differences are unequal, the list is not an AP.
(x) 2a−a=a, 3a−2a=a and 4a−3a=a.
Therefore it is an AP with d=a.
The next terms are 4a+a=5a, 5a+a=6a and 6a+a=7a.
(xi) a²−a, a³−a²=a²(a−1) and a⁴−a³=a³(a−1) are not equal for a general value of a.
Hence the list is not an AP.
(xii) First simplify √8=2√2, √18=3√2 and √32=4√2.
The list becomes √2,2√2,3√2,4√2,….
2√2−√2=√2, 3√2−2√2=√2 and 4√2−3√2=√2.
Therefore it is an AP with d=√2.
The next terms are 5√2=√50, 6√2=√72 and 7√2=√98.
(xiii) √6−√3, √9−√6=3−√6 and √12−√9=2√3−3 are unequal.
Hence the list is not an AP.
(xiv) First write the values: 1²=1, 3²=9, 5²=25 and 7²=49.
The differences are 9−1=8, 25−9=16 and 49−25=24.
Since they are unequal, the list is not an AP.
(xv) First write the values: 1²=1, 5²=25 and 7²=49.
The list is 1,25,49,73,….
25−1=24, 49−25=24 and 73−49=24.
Therefore it is an AP with d=24.
The next terms are 73+24=97, 97+24=121 and 121+24=145.
Final answer
(ii) d=1/2, next terms 4,9/2,5; (iii) d=−2, next terms −9.2,−11.2,−13.2; (iv) d=4, next terms 6,10,14; (v) d=√2, next terms 3+4√2,3+5√2,3+6√2; (vii) d=−4, next terms −16,−20,−24; (viii) d=0, next terms −1/2,−1/2,−1/2; (x) d=a, next terms 5a,6a,7a; (xii) d=√2, next terms √50,√72,√98; (xv) d=24, next terms 97,121,145. Parts (i),(vi),(ix),(xi),(xiii),(xiv) are not APs.