Mathematics · Arithmetic Progressions
NCERT Examples
All 16 worked examples from Arithmetic Progressions, with independently verified solutions and direct navigation.
Mathematics · Chapter 5 · NCERT Examples
NCERT Example
Example 1
For the AP 3/2, 1/2, −1/2, −3/2, …, write the first term and common difference.
Solution
Board-exam working:
The first term is the first listed number, so a=3/2.
d=a₂−a₁=1/2−3/2=(1−3)/2=−2/2=−1.
Check: a₃−a₂=−1/2−1/2=−1 and a₄−a₃=−3/2−(−1/2)=−1.
The common difference is constant.
Final answer
a = 3/2, d = −1.
NCERT Example
Example 2
Which lists form an AP, and for each AP write the next two terms? (i) 4,10,16,22,… (ii) 1,−1,−3,−5,… (iii) −2,2,−2,2,… (iv) 1,1,1,2,2,2,3,3,3,…
Solution
Board-exam working:
Differences are: (i) 6,6,6
(ii) −2,−2,−2
(iii) 4,−4
(iv) 0,0,1.
Thus only (i) and (ii) are APs.
Final answer
(i) AP, d=6, next 28,34; (ii) AP, d=−2, next −7,−9; (iii),(iv) not APs.
NCERT Example
Example 3
Find the 10th term of 2, 7, 12, …
Solution
Board-exam working:
Here a=2 and d=7−2=5.
We need n=10.
a₁₀=a+(10−1)d=2+9×5=2+45=47.
Check: a₉=2+8×5=42 and a₁₁=2+10×5=52, so the difference on both sides is 5.
Final answer
47.
NCERT Example
Example 4
Which term of 21,18,15,… is −81? Is any term 0?
Solution
Board-exam working:
Here a=21 and d=18−21=−3.
For −81: −81=21+(n−1)(−3)=21−3n+3=24−3n.
Thus 3n=105 and n=35.
Check: a₃₅=21+34(−3)=21−102=−81.
For 0: 0=21+(n−1)(−3)=24−3n.
Thus 3n=24 and n=8.
Check: a₈=21+7(−3)=0.
Final answer
−81 is the 35th term; 0 is the 8th term.
NCERT Example
Example 5
Determine the AP whose 3rd term is 5 and 7th term is 9.
Solution
Board-exam working:
a₃=a+2d=5 and a₇=a+6d=9.
Subtract the first equation from the second: (a+6d)−(a+2d)=9−5, so 4d=4 and d=1.
Substitute d=1 in a+2d=5: a+2=5, so a=3.
Therefore the AP begins 3,4,5,6,7,…
its third and seventh terms are 5 and 9.
Final answer
3,4,5,6,7,…
NCERT Example
Example 6
Check whether 301 is a term of 5,11,17,23,…
Solution
Board-exam working:
Here a=5 and d=11−5=6.
Put aₙ=301.
301=5+(n−1)6=5+6n−6=6n−1.
Thus 6n=302 and n=302/6=151/3.
Since 151/3 is not a positive integer, it cannot be a term number.
Final answer
301 is not a term.
NCERT Example
Example 7
How many two-digit numbers are divisible by 3?
Solution
Board-exam working:
The first two-digit multiple of 3 is 12 and the last is 99, so the AP is 12,15,…,99 with a=12,d=3,l=99.
Use l=a+(n−1)d: 99=12+3(n−1).
Thus 87=3(n−1), n−1=29 and n=30.
Check: a₃₀=12+29×3=12+87=99.
Final answer
30 numbers.
NCERT Example
Example 8
Find the 11th term from the end of 10,7,4,…,−62.
Solution
Board-exam working:
Here a=10,d=−3,l=−62.
First find the number of terms: −62=10+(n−1)(−3)=13−3n.
Thus 3n=75 and n=25.
The 11th term from the end is term number 25−11+1=15 from the start.
a₁₅=10+(15−1)(−3)=10−42=−32.
Check from the end: −62+(11−1)3=−62+30=−32.
Final answer
−32.
NCERT Example
Example 9
₹1000 is invested at 8% simple interest yearly. Do yearly accumulated interests form an AP, and what is the interest after 30 years?
Solution
Board-exam working:
Simple interest for one year is PR/100=1000×8/100=₹80.
Therefore the accumulated interests are ₹80,₹160,₹240,… and each consecutive difference is ₹80, so they form an AP with a=80,d=80.
For 30 years, a₃₀=80+(30−1)80=80+29×80=80+2320=₹2400.
Check directly: PRT/100=1000×8×30/100=₹2400.
Final answer
Yes; d=₹80 and the 30-year interest is ₹2400.
NCERT Example
Example 10
Rows of a flower bed contain 23,21,19,…,5 rose plants. Find the number of rows.
Solution
Board-exam working:
The row counts form an AP with a=23,d=21−23=−2 and last term l=5.
Use l=a+(n−1)d: 5=23+(n−1)(−2)=23−2n+2=25−2n.
Thus 2n=20 and n=10.
Check: a₁₀=23+9(−2)=23−18=5.
Final answer
10 rows.
NCERT Example
Example 11
Find the sum of the first 22 terms of 8,3,−2,…
Solution
Board-exam working:
Here a=8,d=3−8=−5,n=22.
S₂₂=22/2[2×8+(22−1)(−5)]=11[16−105]=11(−89)=−979.
The last term is a₂₂=8+21(−5)=−97, and 22/2(8−97)=11(−89)=−979, confirming the sum.
Final answer
−979.
NCERT Example
Example 12
The sum of the first 14 terms is 1050 and the first term is 10. Find the 20th term.
Solution
Board-exam working:
Given a=10,n=14,S₁₄=1050.
Use Sₙ=n/2[2a+(n−1)d].
1050=14/2[2×10+13d]=7(20+13d)=140+91d.
Thus 91d=910 and d=10.
Now a₂₀=a+19d=10+19×10=10+190=200.
Final answer
200.
NCERT Example
Example 13
How many terms of 24,21,18,… must be taken so their sum is 78?
Solution
Board-exam working:
Here a=24,d=−3,Sₙ=78.
78=n/2[2×24+(n−1)(−3)]=n/2[48−3n+3]=n/2(51−3n).
Therefore 156=51n−3n², so 3n²−51n+156=0.
Dividing by 3 gives n²−17n+52=0=(n−4)(n−13).
Hence n=4 or n=13.
Check: S₄=4/2[48−9]=2×39=78, and S₁₃=13/2[48−36]=13/2×12=78.
Final answer
4 terms or 13 terms.
NCERT Example
Example 14
Find (i) the sum of the first 1000 positive integers and (ii) the sum of the first n positive integers.
Solution
Board-exam working:
The positive integers form an AP with a=1,d=1.
Use Sₙ=n(a+l)/2.
(i) Here n=1000 and l=1000.
S₁₀₀₀=1000/2(1+1000)=500×1001=500500.
(ii) For the first n positive integers, l=n.
Therefore Sₙ=n/2(1+n)=n(n+1)/2.
Final answer
(i) 500500; (ii) n(n+1)/2.
NCERT Example
Example 15
Find the sum of the first 24 terms when aₙ=3+2n.
Solution
Board-exam working:
Put n=1,2,3: a₁=3+2=5,a₂=3+4=7,a₃=3+6=9.
Thus a=5 and d=2.
S₂₄=24/2[2×5+(24−1)2]=12[10+46]=12×56=672.
Check using the last term: a₂₄=3+2×24=51, so S₂₄=24/2(5+51)=12×56=672.
Final answer
672.
NCERT Example
Example 16
TV production was 600 sets in year 3 and 700 in year 7, increasing uniformly. Find year 1, year 10, and total for years 1–7.
Solution
Board-exam working:
Let first-year production be a and annual increase be d.
Then a₃=a+2d=600 and a₇=a+6d=700.
Subtracting gives 4d=100, so d=25.
Substituting in a+2d=600 gives a+50=600, hence a=550 sets.
The tenth-year production is a₁₀=a+9d=550+9×25=550+225=775 sets.
The total for the first seven years is S₇=7/2[2×550+(7−1)25]=7/2[1100+150]=7/2×1250=4375 sets.
Check: a₇=550+6×25=700, as given.
Final answer
550 sets; 775 sets; 4375 sets.