Q.The sum of n terms of two arithmetic progressions are in the ratio (3n + 8) : (7n + 15). Find the ratio of their 15th terms.
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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 |
For two APs, the ratio of nth terms can be obtained from the ratio of sums by substituting (2n−1) in place of n; applying this to the given sum ratio at n=15 g …
To get the ratio of nth terms from a ratio of sums, substitute (2n−1) in place of n in the given sum-ratio formula.
For two APs with sums Sn and Sn′, the nth terms satisfy Tn′Tn=S2n−1′S2n−1 (because S2n−1=(2n−1)×(middle term)=(2n−1)Tn for an AP).
…
Showing the 12 most recent of 17 on this concept.
- CBSE 2025Set ANNUAL1 markMCQQ.The common difference of the A.P. 8,15,22,… is(a) 8(b) 15(c) 22(d) 7
›Reveal solutionSolution
The common difference d=7.
For an A.P., d=a2−a1 (equivalently a3−a2, etc.).
…
- CBSE 2025Set ANNUAL1 markMCQQ.The arithmetic mean of two numbers 12 and 15 is(a) 13(b) 13.5(c) 14.5(d) 14
›Reveal solutionSolution
The arithmetic mean of 12 and 15 is 13.5.
For two numbers a and b, the arithmetic mean is 2a+b.
…
- CBSE 2025Set ANNUAL1 markMCQQ.The 15th term of the A.P. 5,13,21,29,… is(a) 113(b) 115(c) 116(d) 117
›Reveal solutionSolution
The 15th term is 117.
For the A.P. 5,13,21,29,…: first term a=5, common difference d=13−5=8.
…
- CBSE 2025Set ANNUAL1 markMCQQ.Which term of the A.P. 9,14,19,24,29,… is 379?(a) 80th(b) 78th(c) 75th(d) 60th
›Reveal solutionSolution
Using an=a+(n−1)d with a=9, d=5, and an=379 gives n=75.
The A.P. is 9,14,19,24,29,… with first term a=9 and common difference d=14−9=5.
an=a+(n−1)d …
- CBSE 2025Set ANNUAL1 markMCQQ.What will be the sum of all natural numbers between 100 and 1000 which are divisible by 5?(a) 94850(b) 98450(c) 99450(d) 94950
›Reveal solutionSolution
The multiples of 5 strictly between 100 and 1000 form an A.P. from 105 to 995 with common difference 5; summing this A.P. gives 98450.
"Between 100 and 1000" is taken to exclude the endpoints themselves, so the sequence of multiples of 5 runs from 105 to 995:
105,110,115,…,995
Number of terms: …
- CBSE 2024Set ANNUAL1 markMCQQ.Find the sum of 17 terms of the A. P. 5, 9, 13, 17, ........:(a) 629(b) 529(c) 615(d) 0
›Reveal solutionSolution
S17=217(10+64)=629.
AP: 5,9,13,17,… has a=5, d=4.
Sn=2n[2a+(n−1)d]
…
- CBSE 2023Set ANNUAL1 markMCQQ.The 10th term of a sequence whose 7th and 12th terms are 34 and 64 respectively is(a) 42(b) 52(c) 63(d) 36
›Reveal solutionSolution
Treating this as an arithmetic sequence, the common difference is 6, giving the 10th term as 52.
Let the sequence be arithmetic with first term a and common difference d. We're told:
a7=34,a12=64
…
- CBSE 2023Set ANNUAL1 markMCQQ.The sum of odd integers from 1 to 100 is:(a) 2525(b) 5050(c) 2500(d) None of these
›Reveal solutionSolution
The odd numbers from 1 to 99 form an AP with 50 terms; their sum is n2.
The odd integers between 1 and 100 are 1,3,5,…,99 — an AP with first term a=1, common difference d=2. The last term is 99, so the number of terms is:
n=299−1+1=50 …
- CBSE 2023Set ANNUAL1 markMCQQ.The sum of n A.M.'s between a and b is equal to(a) 2n(a+b)(b) 2n+1(a+b)(c) n+1(d) 2n+1
›Reveal solutionSolution
Sum of n AMs =2n(a+b); option (a).
Inserting n arithmetic means between a and b gives an AP with n+2 terms. The n means themselves have the same average as a and b, namely 2a+b (NCERT Class 11 Sequences a …
- CBSE 2022Set ANNUAL1 markQ.If nth term of an A.P. is 5n + 1, then its common difference is ............ .
›Reveal solutionSolution
For an=5n+1, consecutive terms differ by the coefficient of n, which is 5.
Given an=5n+1.
…
- CBSE 2022Set ANNUAL1 markQ.If 5th term of an A.P. is 10 and 10th term is 5, then its 15th term is ............ .
›Reveal solutionSolution
Solving for a and d from the given terms gives a=14, d=−1, so the 15th term is 0.
Let the first term be a and common difference d. The nth term is an=a+(n−1)d.
5th term: a+4d=10
10th term: a+9d=5
Subtracting: (a+9d)−(a+4d)=5−10⇒5d=−5⇒d=−1
…
- CBSE 2022Set ANNUAL1 markMCQQ.Fill in the blank with the correct option: the value of the 5th term is ____ in the A.P. 2,8,18,…(a) 52(b) 8(c) 72
›Reveal solutionSolution
Rewriting each term as a multiple of 2 reveals an A.P., whose 5th term is 52.
2=2
8=4×2=22
18=9×2=32
…
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