Exercise 5.1

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.

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.

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.

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.