NCERT Mathematics · Class 10 · English · 2026-27
Arithmetic Progressions — Exercise 5.1
Independently written, step-by-step solutions. Prompts below are concise paraphrases; use the official textbook for the complete question wording.
Choose a question
Question 1
Decide whether the taxi fares, remaining air, per-metre digging charges and compound-interest balances form APs.
Step-by-step solution
(i)
Taxi totals are ₹15, ₹23, ₹31, ₹39, … . Each additional kilometre adds ₹8, so this is an AP with a=15, d=8.
(ii)
If initial air is V, after successive operations it is 3V/4, 9V/16, 27V/64, … . Differences are −3V/16, −9V/64, … and are unequal for V>0. It is not an AP.
(iii)
Charges for successive metres are ₹150, ₹200, ₹250, … . They form an AP with d=50. These are charges per metre; cumulative well costs 150,350,600,… do not form an AP.
(iv)
Balances after successive years are ₹10800, ₹11664, ₹12597.12, … . Differences ₹864 and ₹933.12 differ, so compound-interest balances are not an AP.
Resource ID: CB-10-M-CH5-EX5.1-Q1
Question 2
Write the first four terms for each given first-term/common-difference pair.
Step-by-step solution
Use a, a+d, a+2d, a+3d.
(i)
a=10,d=10: 10,20,30,40.
(ii)
a=−2,d=0: −2,−2,−2,−2.
(iii)
a=4,d=−3: 4,1,−2,−5.
(iv)
a=−1,d=1/2: −1,−1/2,0,1/2.
(v)
a=−1.25,d=−0.25: −1.25,−1.50,−1.75,−2.00.
Resource ID: CB-10-M-CH5-EX5.1-Q2
Question 3
Identify a and d for the four listed APs.
Step-by-step solution
Take a as the first term and subtract the first term from the second to obtain d.
(i)
3,1,−1,−3,…: a=3; d=1−3=−2.
(ii)
−5,−1,3,7,…: a=−5; d=−1−(−5)=4.
(iii)
1/3,5/3,9/3,13/3,…: a=1/3; d=5/3−1/3=4/3.
(iv)
0.6,1.7,2.8,3.9,…: a=0.6; d=1.7−0.6=1.1.
Resource ID: CB-10-M-CH5-EX5.1-Q3
Question 4
Test all fifteen sequences in Question 4; give d and the next three terms for each AP.
Step-by-step solution
Successive differences must all be equal. The answers refer to the textbook’s part numbering.
(i)
2,4,8,16,…: not AP; differences 2,4,8 differ.
(ii)
2,5/2,3,7/2,…: d=1/2; next 4,9/2,5.
(iii)
−1.2,−3.2,−5.2,−7.2,…: d=−2; next −9.2,−11.2,−13.2.
(iv)
−10,−6,−2,2,…: d=4; next 6,10,14.
(v)
3,3+√2,3+2√2,3+3√2,…: d=√2; next 3+4√2,3+5√2,3+6√2.
(vi)
0.2,0.22,0.222,0.2222,…: not AP; the increments 0.02 and 0.002 differ.
(vii)
0,−4,−8,−12,…: d=−4; next −16,−20,−24.
(viii)
−1/2,−1/2,−1/2,−1/2,…: d=0; next three terms are all −1/2.
(ix)
1,3,9,27,…: not AP; differences 2 and 6 differ.
(x)
a,2a,3a,4a,…: d=a; next 5a,6a,7a.
(xi)
a,a²,a³,a⁴,…: not an AP for general a, since a²−a and a³−a² differ. Exceptional values a=0 or a=1 give constant APs with d=0.
(xii)
√2,√8,√18,√32,… simplify to √2,2√2,3√2,4√2,… . Thus d=√2; next 5√2,6√2,7√2.
(xiii)
√3,√6,√9,√12,…: not AP. If the first two differences were equal, √3+3=2√6. Squaring would give 12+6√3=24, which would imply √3=2, a contradiction.
(xiv)
1²,3²,5²,7²,… are 1,9,25,49,…: differences 8 and 16 differ; not AP.
(xv)
1²,5²,7²,73,… are 1,25,49,73,…: d=24; next 97,121,145.
Resource ID: CB-10-M-CH5-EX5.1-Q4