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Mathematics · Arithmetic Progressions · NCERT Exercises

Exercise 5.3

Complete, independently verified solutions for NCERT Exercise 5.3.

Mathematics · Chapter 5 · NCERT Exercises

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Question

Question 1

sum of AP

Find sums: (i) 2,7,12,… 10 terms (ii) −37,−33,−29,… 12 terms (iii) 0.6,1.7,2.8,… 100 terms (iv) 1/15,1/12,1/10,… 11 terms.

Solution

Board-exam working:

Sn=n2[2a+(n−1)d]=n2(a+l)S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)

Use Sₙ=n/2[2a+(n−1)d].

(i) a=2,d=7−2=5,n=10.

S₁₀=10/2[2×2+9×5]=5[4+45]=5×49=245.

(ii) a=−37,d=−33−(−37)=4,n=12.

S₁₂=12/2[2(−37)+11×4]=6[−74+44]=6(−30)=−180.

(iii) a=0.6,d=1.7−0.6=1.1,n=100.

S₁₀₀=100/2[2×0.6+99×1.1]=50[1.2+108.9]=50×110.1=5505.

(iv) a=1/15,d=1/12−1/15=(5−4)/60=1/60,n=11.

S₁₁=11/2[2/15+10/60]=11/2[8/60+10/60]=11/2×3/10=33/20.

Final answer

(i)245; (ii)−180; (iii)5505; (iv)33/20.

Question

Question 2

sum of AP

Find sums: (i) 7+10½+14+…+84 (ii) 34+32+30+…+10 (iii) −5−8−11−…−230.

Solution

Board-exam working:

Sn=n2[2a+(n−1)d]=n2(a+l)S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)

First find n from l=a+(n−1)d, then use Sₙ=n(a+l)/2.

(i) a=7,d=10½−7=7/2,l=84.

Thus 84=7+(n−1)7/2

77×2=7(n−1), so n−1=22 and n=23.

S₂₃=23/2(7+84)=23×91/2=2093/2.

(ii) a=34,d=−2,l=10.

Thus 10=34−2(n−1)

2(n−1)=24, so n=13.

S₁₃=13/2(34+10)=13/2×44=286.

(iii) a=−5,d=−3,l=−230.

Thus −230=−5−3(n−1)

−225=−3(n−1), so n−1=75 and n=76.

S₇₆=76/2(−5−230)=38(−235)=−8930.

Final answer

(i)2093/2; (ii)286; (iii)−8930.

Question

Question 3

sum of AP · table

In an AP, solve: (i) a=5,d=3,aₙ=50, find n and Sₙ; (ii) a=7,a₁₃=35, find d and S₁₃; (iii) a=37,d=3, find a₁₂ and S₁₂; (iv) a=15,S₁₀=125, find d and a₁₀; (v) d=5,S₉=75, find a and a₉; (vi) a=2,d=8,Sₙ=90, find n and aₙ; (vii) a=8,aₙ=62,Sₙ=210, find n and d; (viii) a=−4,d=2,Sₙ=−14, find n and aₙ; (ix) a=3,n=8,S₈=192, find d; (x) l=28,S₉=144, find a.

Solution

Board-exam working:

Sn=n2[2a+(n−1)d]=n2(a+l)S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)

Use aₙ=a+(n−1)d and Sₙ=n(a+aₙ)/2.

(i) 50=5+(n−1)3 gives 45=3(n−1), so n=16.

Then S₁₆=16/2(5+50)=8×55=440.

(ii) 35=7+(13−1)d gives 28=12d, so d=7/3.

Then S₁₃=13/2(7+35)=13×21=273.

(iii) a₁₂=37+(12−1)3=37+33=70.

Then S₁₂=12/2(37+70)=6×107=642.

(iv) Here a=15,n=10,S₁₀=125.

Therefore 125=10/2[2×15+9d]=5(30+9d).

Thus 25=30+9d, so d=−5/9.

Now a₁₀=15+9(−5/9)=10.

(v) Here d=5,n=9,S₉=75.

Therefore 75=9/2[2a+8×5].

Thus 150/9=2a+40, so 2a=−70/3 and a=−35/3.

Now a₉=−35/3+8×5=85/3.

(vi) 90=n/2[2×2+(n−1)8]=n(4n−2).

Thus 2n²−n−45=0=(2n+9)(n−5), so n=5.

Then a₅=2+4×8=34.

(vii) 210=n/2(8+62)=35n, so n=6.

Also 62=8+(6−1)d, hence 54=5d and d=54/5.

(viii) −14=n/2[2(−4)+(n−1)2]=n(n−5).

Thus n²−5n+14=0.

Its discriminant is 25−56=−31, so the printed data has no real or positive-integer n.

(ix) Here a=3,n=8,S₈=192.

Therefore 192=8/2[2×3+7d]=4(6+7d).

Thus 48=6+7d, so 42=7d and d=6.

(x) With l=28,S=144,n=9: 144=9/2(a+28), so 288=9(a+28), a+28=32 and a=4.

Final answer

(i) n=16,Sₙ=440; (ii) d=7/3,S₁₃=273; (iii) a₁₂=70,S₁₂=642; (iv) d=−5/9,a₁₀=10; (v) a=−35/3,a₉=85/3; (vi) n=5,aₙ=34; (vii) n=6,d=54/5; (viii) no real/integer solution for the printed data; (ix) d=6; (x) a=4.

Question

Question 4

sum of AP

How many terms of 9,17,25,… give sum 636?

Solution

Board-exam working:

Sn=n2[2a+(n−1)d]=n2(a+l)S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)

Here a=9,d=8,Sₙ=636.

636=n/2[2×9+(n−1)8]=n/2[18+8n−8]=n/2(8n+10)=n(4n+5).

Thus 4n²+5n−636=0=(4n+53)(n−12).

Therefore n=12 or n=−53/4.

Reject the negative value.

Check: S₁₂=12/2[18+11×8]=6(18+88)=6×106=636.

Final answer

12 terms.

Question

Question 5

sum of AP

First term 5, last 45, sum 400: find n and d.

Solution

Board-exam working:

Sn=n2[2a+(n−1)d]=n2(a+l)S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)

Given a=5,l=45,Sₙ=400.

Use Sₙ=n(a+l)/2.

400=n/2(5+45)=25n, so n=400÷25=16.

Now use l=a+(n−1)d: 45=5+(16−1)d=5+15d.

Thus 40=15d and d=40/15=8/3.

Check: S₁₆=16/2(50)=400.

Final answer

n=16,d=8/3.

Question

Question 6

sum of AP

First and last terms are 17 and 350, d=9. Find n and sum.

Solution

Board-exam working:

Sn=n2[2a+(n−1)d]=n2(a+l)S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)

Given a=17,d=9,l=350.

Use l=a+(n−1)d.

350=17+9(n−1), so 333=9(n−1), n−1=37 and n=38.

S₃₈=38/2(17+350)=19×367=6973.

Check: a₃₈=17+37×9=17+333=350.

Final answer

38 terms; sum 6973.

Question

Question 7

sum of AP

Find S₂₂ when d=7 and a₂₂=149.

Solution

Board-exam working:

Sn=n2[2a+(n−1)d]=n2(a+l)S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)

Use a₂₂=a+21d.

Therefore 149=a+21×7=a+147, so a=2.

Now S₂₂=22/2(a+a₂₂)=11(2+149)=11×151=1661.

Check with the other formula: 22/2[2×2+21×7]=11(4+147)=1661.

Final answer

1661.

Question

Question 8

sum of AP

Find S₅₁ when second and third terms are 14 and 18.

Solution

Board-exam working:

Sn=n2[2a+(n−1)d]=n2(a+l)S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)

a₂=a+d=14 and a₃=a+2d=18.

Subtracting gives d=4.

Then a=14−4=10.

a₅₁=a+50d=10+50×4=210.

S₅₁=51/2(a+a₅₁)=51/2(10+210)=51/2×220=51×110=5610.

Final answer

5610.

Question

Question 9

sum of AP

S₇=49 and S₁₇=289. Find Sₙ.

Solution

Board-exam working:

Sn=n2[2a+(n−1)d]=n2(a+l)S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)

S₇=7/2[2a+6d]=7(a+3d)=49, so a+3d=7.

S₁₇=17/2[2a+16d]=17(a+8d)=289, so a+8d=17.

Subtracting gives 5d=10, so d=2.

Then a+3×2=7 gives a=1.

Therefore Sₙ=n/2[2×1+(n−1)2]=n/2(2n)=n².

Check: S₇=49 and S₁₇=289.

Final answer

Sₙ=n².

Question

Question 10

sum of AP

Show sequences aₙ=3+4n and aₙ=9−5n are APs; find each S₁₅.

Solution

Board-exam working:

Sn=n2[2a+(n−1)d]=n2(a+l)S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)

(i) a₁=3+4=7,a₂=3+8=11,a₃=3+12=15.

The differences are 11−7=4 and 15−11=4, so it is an AP with a=7,d=4.

Then S₁₅=15/2[2×7+14×4]=15/2(14+56)=15/2×70=525.

(ii) a₁=9−5=4,a₂=9−10=−1,a₃=9−15=−6.

The differences are −1−4=−5 and −6−(−1)=−5, so it is an AP with a=4,d=−5.

Then S₁₅=15/2[2×4+14(−5)]=15/2(8−70)=15/2(−62)=−465.

Final answer

(i) AP,d=4,S₁₅=525; (ii) AP,d=−5,S₁₅=−465.

Question

Question 11

sum to nth term

If Sₙ=4n−n², find S₁, S₂, a₂, a₃, a₁₀ and aₙ.

Solution

Board-exam working:

Sn=n2[2a+(n−1)d]=n2(a+l)S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)

S₁=4(1)−1²=4−1=3 and S₂=4(2)−2²=8−4=4.

a₂=S₂−S₁=4−3=1.

Also S₃=4(3)−3²=12−9=3, so a₃=S₃−S₂=3−4=−1.

In general, aₙ=Sₙ−Sₙ₋₁=[4n−n²]−[4(n−1)−(n−1)²].

Simplifying gives aₙ=5−2n.

Therefore a₁₀=5−2(10)=5−20=−15.

Final answer

3;4;1;−1;−15; aₙ=5−2n.

Question

Question 12

application · sum of AP

Find the sum of first 40 positive integers divisible by 6.

Solution

Board-exam working:

Sn=n2[2a+(n−1)d]=n2(a+l)S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)

The numbers are 6,12,18,…, so a=6,d=6,n=40.

The 40th term is a₄₀=6+(40−1)6=6+234=240.

S₄₀=40/2(6+240)=20×246=4920.

Final answer

4920.

Question

Question 13

application · sum of AP

Find the sum of first 15 multiples of 8.

Solution

Board-exam working:

Sn=n2[2a+(n−1)d]=n2(a+l)S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)

The multiples are 8,16,24,…, so a=8,d=8,n=15.

The 15th term is a₁₅=8+(15−1)8=8+112=120.

S₁₅=15/2(8+120)=15/2×128=15×64=960.

Final answer

960.

Question

Question 14

sum of AP

Find the sum of odd numbers between 0 and 50.

Solution

Board-exam working:

Sn=n2[2a+(n−1)d]=n2(a+l)S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)

The odd numbers are 1,3,5,…,49, so a=1,d=2,l=49.

Find n: 49=1+(n−1)2, so 48=2(n−1), n−1=24 and n=25.

S₂₅=25/2(1+49)=25/2×50=25×25=625.

Final answer

625.

Question

Question 15

application · sum of AP

Daily delay penalty is ₹200,₹250,₹300,… Find total for 30 days.

Solution

Board-exam working:

Sn=n2[2a+(n−1)d]=n2(a+l)S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)

The daily penalties form an AP with a=₹200,d=₹50,n=30.

S₃₀=30/2[2×200+(30−1)50]=15[400+1450]=15×1850=₹27,750.

Check: the 30th-day penalty is 200+29×50=₹1650, and 30/2(200+1650)=₹27,750.

Final answer

₹27,750.

Question

Question 16

application · sum of AP

₹700 funds seven prizes, each ₹20 less than the previous. Find the prizes.

Solution

Board-exam working:

Sn=n2[2a+(n−1)d]=n2(a+l)S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)

Let the first prize be ₹a.

There are n=7 prizes and d=−₹20.

700=7/2[2a+(7−1)(−20)]=7/2(2a−120).

Multiplying by 2/7 gives 200=2a−120, so 2a=320 and a=160.

Subtracting ₹20 successively gives ₹160,₹140,₹120,₹100,₹80,₹60,₹40.

Their sum is ₹700.

Final answer

₹160,₹140,₹120,₹100,₹80,₹60,₹40.

Question

Question 17

application · sum of AP

Three sections of each Class I–XII plant the class number of trees. Find total.

Solution

Board-exam working:

Sn=n2[2a+(n−1)d]=n2(a+l)S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)

For one section, the numbers planted by Classes I–XII are 1,2,3,…,12.

This is an AP with a=1,l=12,n=12.

One-section total=12/2(1+12)=6×13=78 trees.

There are three sections, so total=3×78=234 trees.

Final answer

234 trees.

Question

Question 18

application · sum of AP

A spiral has 13 semicircles of radii 0.5,1.0,1.5,… cm. Find total length using π=22/7.

Spiral made from successive semicircles with centres alternating between A and B.
Figure 5.4: spiral of successive semicircles.

Solution

Board-exam working:

Sn=n2[2a+(n−1)d]=n2(a+l)S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)

The radii form an AP with a=0.5 cm,d=0.5 cm,n=13.

The last radius is l=0.5+12×0.5=6.5 cm.

Sum of radii=13/2(0.5+6.5)=13/2×7=45.5 cm.

The length of a semicircle of radius r is πr.

Therefore total length=π×45.5=22/7×45.5=22×6.5=143 cm.

Final answer

143 cm.

Question

Question 19

application · sum of AP

200 logs are stacked 20 in bottom row, then 19,18,… Find rows and top row count.

Logs stacked in rows with each higher row containing one fewer log.
Figure 5.5: stack of 200 logs.

Solution

Board-exam working:

Sn=n2[2a+(n−1)d]=n2(a+l)S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)

The row counts form an AP with a=20,d=−1,Sₙ=200.

200=n/2[2×20+(n−1)(−1)]=n/2(41−n).

Thus 400=41n−n², so n²−41n+400=0=(n−16)(n−25).

Hence n=16 or n=25.

If n=25, the top row would contain 20+24(−1)=−4 logs, which is impossible.

Therefore n=16.

The top row contains a₁₆=20+15(−1)=5 logs.

Check: 16/2(20+5)=8×25=200.

Final answer

16 rows; 5 logs on top.

Question

Question 20

application · sum of AP

Ten potatoes start 5 m from a bucket and are 3 m apart. Find total out-and-back running distance.

A runner carrying potatoes placed along a straight line back to a bucket.
Figure 5.6: potato race layout.

Solution

Board-exam working:

Sn=n2[2a+(n−1)d]=n2(a+l)S_n=\frac{n}{2}[2a+(n-1)d]=\frac{n}{2}(a+l)

The one-way distances from the bucket are 5,8,11,… with a=5,d=3,n=10.

The last distance is a₁₀=5+(10−1)3=5+27=32 m.

Sum of one-way distances=10/2(5+32)=5×37=185 m.

Each potato requires an outward and return journey, so total distance=2×185=370 m.

Final answer

370 m.