NCERT Exercise 5.2 — Arithmetic Progressions

Question

Question 1

nth term · table

Complete table: (i) a=7,d=3,n=8; (ii) a=−18,n=10,aₙ=0; (iii) d=−3,n=18,aₙ=−5; (iv) a=−18.9,d=2.5,aₙ=3.6; (v) a=3.5,d=0,n=105.

Solution

Board-exam working:

a_n=a+(n-1)d

Use aₙ=a+(n−1)d in every row.

(i) a₈=7+(8−1)×3=7+21=28.

(ii) 0=−18+(10−1)d, so 18=9d and d=18÷9=2.

Check: −18+9×2=0.

(iii) −5=a+(18−1)(−3)=a−51, so a=−5+51=46.

Check: 46−51=−5.

(iv) 3.6=−18.9+(n−1)×2.5, so 22.5=2.5(n−1), n−1=9 and n=10.

Check: −18.9+9×2.5=3.6.

(v) a₁₀₅=3.5+(105−1)×0=3.5.

Final answer

(i) aₙ=28; (ii) d=2; (iii) a=46; (iv) n=10; (v) aₙ=3.5.

Question

Question 2

nth term

Choose and justify: (i) 30th term of 10,7,4,…; (ii) 11th term of −3,−1/2,2,…

Solution

Board-exam working:

a_n=a+(n-1)d

Use aₙ=a+(n−1)d.

(i) a=10,d=7−10=−3,n=30.

Therefore a₃₀=10+29(−3)=10−87=−77, which is option C.

(ii) a=−3,d=−1/2−(−3)=5/2,n=11.

Therefore a₁₁=−3+10×5/2=−3+25=22, which is option B.

Final answer

(i) C, −77; (ii) B, 22.

Question

Question 3

missing terms

Fill missing terms: (i) 2,□,26 (ii) □,13,□,3 (iii) 5,□,□,9 (iv) −4,□,□,□,□,6 (v) □,38,□,□,□,−22.

Solution

Board-exam working:

a_n=a+(n-1)d

Use aₙ=a+(n−1)d and divide the change between known terms equally.

(i) 26=2+2d, so 2d=24 and d=12.

The middle term is 2+12=14.

(ii) 3=13+2d, so 2d=−10 and d=−5.

Hence the first term is 13−(−5)=18 and the third is 13−5=8.

(iii) 9=5+3d, so 3d=4 and d=4/3.

The missing terms are 5+4/3=19/3 and 19/3+4/3=23/3.

(iv) 6=−4+5d, so 5d=10 and d=2.

Adding 2 successively gives −2,0,2,4.

(v) −22=38+4d, so 4d=−60 and d=−15.

The first term is 38−(−15)=53

continuing gives 23,8,−7,−22.

In every completed sequence, each consecutive difference is constant.

Final answer

(i) 14; (ii) 18,8; (iii) 19/3,23/3; (iv) −2,0,2,4; (v) 53,23,8,−7.

Question

Question 4

term number

Which term of 3,8,13,18,… is 78?

Solution

Board-exam working:

a_n=a+(n-1)d

Here a=3,d=8−3=5.

Put aₙ=78.

78=3+(n−1)5=3+5n−5=5n−2.

Thus 5n=80 and n=16.

Check: a₁₆=3+15×5=3+75=78.

Final answer

16th term.

Question

Question 5

number of terms

Find number of terms: (i) 7,13,19,…,205 (ii) 18,15½,13,…,−47.

Solution

Board-exam working:

a_n=a+(n-1)d

Use l=a+(n−1)d.

(i) a=7,d=6,l=205.

Then 205=7+6(n−1), so 198=6(n−1), n−1=33 and n=34.

Check: 7+33×6=205.

(ii) a=18,d=15½−18=−5/2,l=−47.

Then −47=18+(n−1)(−5/2), so −65=−5/2(n−1).

Multiplying by −2/5 gives n−1=26, hence n=27.

Check: 18+26(−5/2)=18−65=−47.

Final answer

(i) 34; (ii) 27.

Question

Question 6

term number

Is −150 a term of 11,8,5,2,…?

Solution

Board-exam working:

a_n=a+(n-1)d

Here a=11 and d=8−11=−3.

Put aₙ=−150 in aₙ=a+(n−1)d.

−150=11+(n−1)(−3)=11−3n+3=14−3n.

Therefore 3n=164 and n=164/3.

Since a term number must be a positive integer, −150 is not a term of this AP.

Final answer

No.

Question

Question 7

nth term

Find the 31st term when the 11th is 38 and 16th is 73.

Solution

Board-exam working:

a_n=a+(n-1)d

a₁₁=a+10d=38 and a₁₆=a+15d=73.

Subtracting gives 5d=35, so d=7.

Substitute in a+10d=38: a+70=38, hence a=−32.

a₃₁=a+30d=−32+30×7=−32+210=178.

Check: a₁₁=−32+70=38 and a₁₆=−32+105=73.

Final answer

178.

Question

Question 8

nth term

An AP has 50 terms, third term 12 and last term 106. Find the 29th term.

Solution

Board-exam working:

a_n=a+(n-1)d

a₃=a+2d=12 and a₅₀=a+49d=106.

Subtracting gives 47d=94, so d=2.

Substitute in a+2d=12: a+4=12, hence a=8.

a₂₉=a+28d=8+28×2=8+56=64.

Check: a₅₀=8+49×2=106.

Final answer

64.

Question

Question 9

term number

Third and ninth terms are 4 and −8. Which term is zero?

Solution

Board-exam working:

a_n=a+(n-1)d

a₃=a+2d=4 and a₉=a+8d=−8.

Subtracting gives 6d=−12, so d=−2.

Substitute in a+2d=4: a−4=4, hence a=8.

Put aₙ=0: 0=8+(n−1)(−2)=10−2n.

Thus 2n=10 and n=5.

Check: a₅=8+4(−2)=0.

Final answer

5th term.

Question

Question 10

common difference

The 17th term exceeds the 10th by 7. Find d.

Solution

Board-exam working:

Write a₁₇=a+16d and a₁₀=a+9d.

The given condition is a₁₇−a₁₀=7.

Therefore (a+16d)−(a+9d)=7.

Cancelling a gives 7d=7, so d=7÷7=1.

Final answer

d=1.

Question

Question 11

term number

Which term of 3,15,27,39,… is 132 more than its 54th term?

Solution

Board-exam working:

a_n=a+(n-1)d

Here a=3 and d=15−3=12.

First find a₅₄=3+(54−1)12=3+53×12=3+636=639.

The required term is 639+132=771.

Put aₙ=771: 771=3+(n−1)12.

Thus 768=12(n−1), n−1=64 and n=65.

Check: a₆₅=3+64×12=771.

Final answer

65th term.

Question

Question 12

nth term

Two APs have the same d. Their 100th terms differ by 100. Find the difference of their 1000th terms.

Solution

Board-exam working:

a_n=a+(n-1)d

Let the first terms be a and b and let the common difference of both APs be d.

Their 100th-term difference is [a+99d]−[b+99d]=a−b=100.

Their 1000th-term difference is [a+999d]−[b+999d]=a−b.

Since a−b=100, this difference is also 100.

Final answer

100.

Question

Question 13

application · number of terms

How many three-digit numbers are divisible by 7?

Solution

Board-exam working:

a_n=a+(n-1)d

The first three-digit multiple of 7 is 105 and the last is 994, so the AP is 105,112,…,994 with a=105,d=7,l=994.

Use l=a+(n−1)d: 994=105+7(n−1).

Thus 889=7(n−1), n−1=127 and n=128.

Check: a₁₂₈=105+127×7=105+889=994.

Final answer

128.

Question

Question 14

application · number of terms

How many multiples of 4 lie between 10 and 250?

Solution

Board-exam working:

a_n=a+(n-1)d

The first multiple of 4 greater than 10 is 12 and the last below 250 is 248.

Thus a=12,d=4,l=248.

Use l=a+(n−1)d: 248=12+4(n−1).

Hence 236=4(n−1), n−1=59 and n=60.

Check: a₆₀=12+59×4=12+236=248.

Final answer

60.

Question

Question 15

nth term

For what n are terms of 63,65,67,… and 3,10,17,… equal?

Solution

Board-exam working:

a_n=a+(n-1)d

For the first AP, a=63,d=2, so aₙ=63+2(n−1)=61+2n.

For the second AP, a=3,d=7, so aₙ=3+7(n−1)=7n−4.

Equate them: 61+2n=7n−4.

Thus 65=5n and n=13.

Check: 63+12×2=87 and 3+12×7=87.

Final answer

n=13.

Question

Question 16

forming an AP

Determine the AP whose third term is 16 and whose 7th exceeds its 5th by 12.

Solution

Board-exam working:

a_n=a+(n-1)d

The third term gives a+2d=16.

Also a₇−a₅=12.

Therefore (a+6d)−(a+4d)=12, so 2d=12 and d=6.

Substitute in a+2d=16: a+12=16, hence a=4.

The AP is 4,10,16,22,… Check: a₇=40 and a₅=28, whose difference is 12.

Final answer

4,10,16,22,…

Question

Question 17

term from end

Find the 20th term from the end of 3,8,13,…,253.

Solution

Board-exam working:

a_n=a+(n-1)d

Here a=3,d=5,l=253.

First find the total terms: 253=3+5(n−1), so 250=5(n−1), n−1=50 and n=51.

The 20th term from the end is term number 51−20+1=32 from the start.

a₃₂=3+(32−1)5=3+31×5=3+155=158.

Check from the end: 253−19×5=158.

Final answer

158.

Question

Question 18

forming an AP

a₄+a₈=24 and a₆+a₁₀=44. Find the first three terms.

Solution

Board-exam working:

a_n=a+(n-1)d

a₄+a₈=(a+3d)+(a+7d)=2a+10d=24.

a₆+a₁₀=(a+5d)+(a+9d)=2a+14d=44.

Subtracting the first equation from the second gives 4d=20, so d=5.

Substitute in 2a+10d=24: 2a+50=24, so 2a=−26 and a=−13.

The first three terms are a,a+d,a+2d=−13,−8,−3.

Check: a₄=2,a₈=22 and their sum is 24.

Final answer

−13,−8,−3.

Question

Question 19

application · term number

Subba Rao began in 1995 at ₹5000 with ₹200 annual increments. When did salary reach ₹7000?

Solution

Board-exam working:

a_n=a+(n-1)d

The annual salaries form an AP with a=₹5000,d=₹200.

Put aₙ=₹7000.

7000=5000+(n−1)200.

Thus 2000=200(n−1), n−1=10 and n=11.

Taking 1995 as year 1, year 11 is 1995+(11−1)=2005.

Check: 5000+10×200=₹7000.

Final answer

2005.

Question

Question 20

application · term number

Ramkali saves ₹5 in week 1 and increases weekly saving by ₹1.75. In which week is it ₹20.75?

Solution

Board-exam working:

a_n=a+(n-1)d

The weekly savings form an AP with a=₹5,d=₹1.75.

Put aₙ=₹20.75.

20.75=5+(n−1)1.75.

Thus 15.75=1.75(n−1), n−1=15.75÷1.75=9 and n=10.

Check: a₁₀=5+9×1.75=5+15.75=₹20.75.

Final answer

10th week.