Q.A man saved Rs 66000 in 20 years. In each succeeding year after the first year he saved Rs 200 more than what he saved in the previous year. How much did he save in the first year?
Concept understanding — Arithmetic Progression
Arithmetic Progression: The Pattern of Equal Steps
Imagine you're climbing a staircase where every step has the exact same height. If the first step takes you to 3 feet, and each step after that adds exactly 2 feet, your heights would be: 3, 5, 7, 9, 11, ... That's an arithmetic progression — a sequence where you move forward by adding the same number every time.
The Core Idea
An Arithmetic Progression (AP) is a list of numbers where the difference between any two consecutive terms is constant. This constant is called the common difference, usually denoted by d.
If the first term is a, then the sequence looks like:
a, a+d, a+2d, a+3d, a+4d, …
The pattern is simple: you start at a, then keep adding d to get the next term.
The common difference d can be positive, negative, or even zero. If d=0, all terms are the same — that's still an AP, just a boring one.
The General Term (nth term)
What if you want the 100th term without writing all 100 numbers? There's a formula.
The first term is a (think of it as a+0⋅d).
The second term is a+d (that's a+1⋅d).
The third term is a+2d.
Notice the pattern: the term number minus 1 tells you how many times d has been added.
So the nth term (also called the general term) is:
Tn=a+(n−1)d
Tn=a+(n−1)d
Example: For the AP 3, 5, 7, 9, ... we have a=3, d=2.
The 10th term: T10=3+(10−1)⋅2=3+18=21.
Why "Arithmetic"?
The name comes from an old property: in an AP, every term (except the first and last) is the arithmetic mean of its neighbours. For three consecutive terms x,y,z in an AP:
y=2x+z
Check: in 3, 5, 7, we have 5=23+7=5. This works for any three consecutive terms.
Sum of the First n Terms
Sometimes you need the total of the first n terms. There's a clever trick.
Write the sum forwards: Sn=a+(a+d)+(a+2d)+⋯+[a+(n−1)d]
Write it backwards: Sn=[a+(n−1)d]+[a+(n−2)d]+⋯+a
Add them term by term. Each pair adds to 2a+(n−1)d, and there are n such pairs. So:
2Sn=n[2a+(n−1)d]
Therefore:
Sn=2n[2a+(n−1)d]
There's another useful form. Since the last term l=a+(n−1)d, we can write:
Sn=2n(a+l)
This is beautiful: the sum of an AP is just the number of terms times the average of the first and last term.
Example: Sum of first 10 terms of 3, 5, 7, ...
S10=210[2⋅3+(10−1)⋅2]=5[6+18]=5×24=120
Quick Reference
| What you need | Formula |
|---|---|
| nth term | Tn=a+(n−1)d |
| Sum of n terms | Sn=2n[2a+(n−1)d] |
| Sum using last term | Sn=2n(a+l) |
| Common difference | d=Tn+1−Tn |
To check if three numbers p,q,r are in AP, just verify 2q=p+r. If that holds, they're equally spaced.
Common Mistakes to Avoid
- Confusing n with the term value. n is the position (1st, 2nd, 3rd...), not the number itself.
- Forgetting the (n−1) in the nth term. Many students write a+nd by mistake. The first term has zero d's added, so it's a+(1−1)d=a.
- Using the wrong n in the sum formula. If you want the sum of the first 20 terms, n=20, not 21.
A Real-World Feel
APs show up everywhere: monthly rent increasing by a fixed amount each year, the number of seats in each row of an auditorium (if each row has 2 more seats than the previous), or even the simple act of counting by 5s: 5, 10, 15, 20, ... That's an AP with a=5, d=5.
Once you see the pattern of equal steps, you'll spot arithmetic progressions all around you.
Arithmetic Progression is one of the most exam-heavy topics in the NCERT Class 11 Mathematics chapter on Sequences and Series, matching frequent searches for "arithmetic progression nth term and sum formula" or "AP important questions class 11 maths". Its equal-step pattern also shows up regularly in JEE Main and state CET numerical-ability sections, often disguised as real-world word problems like EMIs or seating arrangements.
The key idea is that the yearly savings form an arithmetic progression (AP), and the total saved over 20 years is the sum of this AP.
Let the first year's saving be a (in Rs). The common difference is d=200, and the number of terms is n=20.
The sum of an AP is given by:
Sn=2n[2a+(n−1)d]
Substitute the known values:
66000=220[2a+(20−1)×200]
66000=10[2a+3800]
Divide both sides by 10:
6600=2a+3800
2a=6600−3800=2800
a=1400
He saved 1400 rupees in the first year.
The problem is an arithmetic series where the total saved over 20 years is Rs 66,000, with each year’s saving increasing by Rs 200. Using the sum formula for an arithmetic progression, the first year’s saving is found to be Rs 1,400.
The key here is recognising that the yearly savings form an arithmetic progression (AP). In an AP, the difference between consecutive terms is constant — in this case, Rs 200. The total saved over 20 years is the sum of the first 20 terms of this AP. We are asked for the first term, which is the saving in the first year.
Let’s denote the saving in the first year as a (in rupees). Then:
- Year 1: a
- Year 2: a+200
- Year 3: a+400
- …
- Year 20: a+19×200=a+3800
The sum of the first n terms of an AP is given by:
Sn=2n[2a+(n−1)d]
where n is the number of terms, a is the first term, and d is the common difference.
Here, n=20, d=200, and S20=66000.
- Write the sum formula with the given values:
66000=220[2a+(20−1)×200]
- Simplify the factor outside: 220=10, so:
66000=10[2a+19×200]
- Compute 19×200: 19×200=3800, so:
66000=10[2a+3800]
- Divide both sides by 10:
6600=2a+3800
- Isolate 2a:
2a=6600−3800=2800
- Solve for a:
a=22800=1400
A common mistake is to forget that the increase applies from the second year onward, so the 20th year’s saving is a+19d, not a+20d. Always check: for n terms, the last term is a+(n−1)d.
You can also solve by first finding the average saving per year: total divided by 20 gives Rs 3,300. In an AP, the average of all terms equals the average of the first and last terms. So 2a+(a+3800)=3300, which gives a=1400 — a quicker mental check.
He saved Rs 1,400 in the first year.
- CA Foundation 2026Set may-20261 markMCQQ.The sum of the first n terms of an arithmetic progression (A.P.) is 4n2+3n. The 10th term of the A.P. is ______. (A) 77 (B) 83 (C) 81 (D) 79
›Reveal solutionSolution
Use an=Sn−Sn−1: a10=S10−S9=430−351=79.
Step 1 — Recall the relation
The n-th term of any sequence equals the difference of consecutive partial sums:
an=Sn−Sn−1
Step 2 — Evaluate S10 and S9
S10=4(10)2+3(10)=400+30=430
S9=4(9)2+3(9)=324+27=351
Step 3 — Subtract
a10=430−351=79
TipAlternatively, Sn=4n2+3n has the form of an AP sum with common difference d=8 and first term a1=S1=7, so a10=7+9(8)=79 — same answer.
Watch outDon't just plug n=10 into Sn — that gives the sum of ten terms (430), not the tenth term. You must subtract S9.
✓Final answer(D) 79
- CA Foundation 2026Set may-20261 markMCQQ.If the sum of 4th and 8th term of an arithmetic progression (A.P.) is 120, then the 6th term of the A.P. is ______. (A) 10 (B) 70 (C) 60 (D) 100
›Reveal solutionSolution
Terms equidistant from a middle term average to it: a4+a8=2a6, so a6=120/2=60.
Step 1 — Write the two terms
a4=a+3d,a8=a+7d
Step 2 — Add them
a4+a8=(a+3d)+(a+7d)=2a+10d=2(a+5d)
But a+5d=a6, so:
a4+a8=2a6
Step 3 — Solve
2a6=120 ⇒ a6=60
TipIn an AP, the sum of two terms equidistant from a term equals twice that term (here 4 and 8 are symmetric about 6). This shortcut avoids solving for a and d separately.
Watch outDon't try to find a and d individually — one equation cannot pin down both, and you don't need to. The symmetry gives a6 directly.
✓Final answer(C) 60
- CA Foundation 2025Set jan-20251 markMCQQ.The sum of the 4th and 8th term of an AP is 10. Then the sum of first eleven terms of the series is (A) 33 (B) 22 (C) 44 (D) 55
›Reveal solutionSolution
t4+t8=2a+10d=10, and S11=211(2a+10d)=211(10)=55.
Step 1 — Express the two terms
t4=a+3d,t8=a+7d
t4+t8=2a+10d=10
Step 2 — Write the sum of 11 terms
Sn=2n(2a+(n−1)d)
For n=11: S11=211(2a+10d).
Step 3 — Substitute the known value
S11=211(10)=11×5=55
Why the other options are wrong: 33, 22 and 44 come from using a wrong n or mis-forming 2a+(n−1)d.
Watch out2a+(11−1)d=2a+10d — exactly the quantity you were given. Don't waste time solving for a and d separately; they aren't individually determined.
TipS11=11×t6 (the middle term), and t6=2t4+t8=5, so S11=55 instantly.
✓Final answer(D) 55
- CA Foundation 2025Set jan-20251 markMCQQ.Find the 9th term of the A.P. 8,5,2,−1,−4,…… (A) −10 (B) −24 (C) −16 (D) −4
›Reveal solutionSolution
a=8, d=−3, so t9=8+(9−1)(−3)=−16.
Step 1 — Identify a and d
a=8,d=5−8=−3
Step 2 — Apply the nth-term formula
tn=a+(n−1)d
For n=9:
t9=8+(9−1)(−3)=8+8(−3)
Step 3 — Evaluate
t9=8−24=−16
Why the other options are wrong: (A) −10 uses d=−2; (B) −24 forgets the +8; (D) −4 stops at the 5th term.
Watch outUse (n−1) steps of d, not n. For the 9th term you add d eight times, giving 8×(−3)=−24.
TipWith a negative d, list a couple more terms mentally (−4,−7,−10,−13,−16) as a quick check on t9.
✓Final answer(C) −16
- CA Foundation 2025Set jan-20251 markMCQQ.The sum of series 1+2+3+…… is 55. The number of terms is : (A) 40 (B) 30 (C) 20 (D) 10
›Reveal solutionSolution
2n(n+1)=55⇒n(n+1)=110⇒n=10.
Step 1 — Use the sum of first n natural numbers
Sn=2n(n+1)
Step 2 — Set equal to 55 and simplify
2n(n+1)=55⇒n(n+1)=110
Step 3 — Solve the quadratic
n2+n−110=0⇒(n−10)(n+11)=0
Taking the positive root, n=10.
Why the other options are wrong: 20, 30 and 40 give sums of 210, 465 and 820 respectively — far above 55.
Watch outReject the negative root n=−11: a count of terms must be a positive whole number.
TipRecognise 55 as a triangular number (T10=55) to jump straight to n=10.
✓Final answer(D) 10
- CA Foundation 2025Set may-20251 markMCQQ.Find the sum of n terms of the A.P., whose nth term is 5n+1. (A) 2n (B) 72n (C) 2n(7+5n) (D) 2n(7+4n)
›Reveal solutionSolution
Sn=2n(a1+an)=2n(6+5n+1)=2n(5n+7).
Step 1 — Find the first term
Given an=5n+1, put n=1: a1=5(1)+1=6.
Step 2 — Apply the A.P. sum formula
Sn=2n(a1+an)
Step 3 — Substitute and simplify
Sn=2n(6+(5n+1))=2n(5n+7)=2n(7+5n)
Why the other options are wrong: (D) 2n(7+4n) uses the wrong common-difference term (coefficient 4 instead of 5); (A) and (B) are unrelated single fractions.
Watch outDo not confuse the nth term with the sum — first extract a1 (and, if needed, the common difference d=5) before summing.
TipWhen an is linear in n, Sn=2n(a1+an) is the fastest route — no need to find d explicitly.
✓Final answer(C) 2n(7+5n)
- CA Foundation 2025Set may-20251 markMCQQ.Insert 4 numbers between 2 and 22 such that the resulting sequence is an Arithmetic Progression (A.P.). (A) 4, 8, 12, 16 (B) 5, 9, 13, 17 (C) 4, 10, 15, 19 (D) 6, 10, 14, 18
›Reveal solutionSolution
6 terms from 2 to 22 ⇒ d=20/5=4 ⇒ inserted numbers 6, 10, 14, 18.
Step 1 — Count the terms
Inserting 4 numbers between 2 and 22 gives 4+2=6 terms with a1=2 and a6=22.
Step 2 — Find the common difference
a6=a1+5d⇒22=2+5d⇒d=520=4
Step 3 — Build the sequence
2,6,10,14,18,22
The four inserted (arithmetic mean) numbers are 6,10,14,18.
Why the other options are wrong: (A) 4,8,12,16 uses d=4 but starts from the wrong first inserted value; (B) and (C) do not form a constant-difference sequence ending at 22.
Watch outThere are 5 gaps (not 4) between the 6 terms — divide the total span 20 by 5, not by 4.
TipInserting k arithmetic means between a and b: d=(b−a)/(k+1).
✓Final answer(D) 6, 10, 14, 18
- CA Foundation 2025Set sep-20251 markMCQQ.The common difference of the arithmetic progression 31,31−3b,31−6b,… is __________. (A) −b (B) b (C) −3b (D) 3b
›Reveal solutionSolution
d=t2−t1=31−3b−31=−b.
Step 1 — Recall the definition
d=tk+1−tk(same for every consecutive pair)
Step 2 — Subtract consecutive terms
d=31−3b−31=3(1−3b)−1=3−3b=−b
Step 3 — Confirm with the next pair
31−6b−31−3b=3−3b=−b ✓
The difference is constant, confirming it is a valid AP with d=−b.
Why the other options are wrong: (B) b has the wrong sign; (C) −3b and (D) 3b forget to divide the numerator difference by 3.
Watch outKeep the common denominator 3 through the subtraction — dropping it gives −3b (option C), a classic trap.
TipFor an AP written with a common denominator, just subtract numerators, then divide once.
✓Final answer(A) −b
- CA Foundation 2025Set sep-20251 markMCQQ.The sum of all natural numbers between 200 and 600 those are divisible by 13 is __________. (A) 12493 (B) 14493 (C) 16493 (D) 18493
›Reveal solutionSolution
Multiples of 13 in range: 208 to 598, 31 terms; S=231(208+598)=12,493.
Step 1 — Find the first and last multiples of 13
13×16=208 (first multiple above 200) and 13×46=598 (last multiple below 600).
Step 2 — Count the terms
n=46−16+1=31
Step 3 — Sum the arithmetic series
Sn=2n(a+l)
S31=231(208+598)=231×806=31×403=12,493
Why the other options are wrong: (B) 14,493, (C) 16,493 and (D) 18,493 result from an incorrect term count or including 195/605 outside the (200, 600) range.
Watch out"Between 200 and 600" excludes both endpoints — start at 208, not 195, and stop at 598, not 611.
TipConvert the divisibility count into an AP: index the multiples (13×16 to 13×46) and the term count is just the difference of the indices plus one.
✓Final answer(A) 12493
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