📚 Class 10 CBSE – Mathematics
Topic: Arithmetic Progressions (AP)
Arithmetic Progression – A sequence of numbers where the difference between consecutive terms is constant.
Common difference (denoted (d)) is that constant difference.
1. Simple Explanation
| Item | Details |
|---|---|
| Definition | In an AP, each term after the first is obtained by adding the common difference (d) to the preceding term. |
| Notation | First term (a). (n^{th}) term (T_n). Sequence written as (a,, a+d,, a+2d,, a+3d,\dots) |
| Example | Sequence (3,, 7,, 11,, 15,\dots) has (a=3), (d=4). |
2. Key Points & Formulas
| Concept | Formula | What it gives |
|---|---|---|
| (n^{th}) term | (T_n = a + (n-1)d) | Value of any term |
| Sum of first (n) terms | (S_n = \frac{n}{2},(2a + (n-1)d)) or (S_n = \frac{n}{2},(T_1+T_n)) | Total of the first (n) terms |
| Sum of (n) terms when last term (l) is known | (S_n = \frac{n}{2},(a+l)) | Same as above |
| Number of terms between two given terms | If terms (x) and (y) (where (y>x)) are in the AP, number of terms between them = (\frac{y-x}{d} - 1) | Useful for “between” questions |
| Finding common difference from two terms | (d = \frac{T_m - T_k}{m-k}) | If you know two terms and their positions |
| Finding first term from two terms | (a = T_k - (k-1)d) | If you know a term and its position |
| Arithmetic Mean (AM) | (AM = \frac{a + l}{2}) | Middle term if the number of terms is odd |
3. Worked Example
Problem
In an arithmetic progression, the 5th term is 23 and the 12th term is 56.
- Find the common difference (d).
- Find the first term (a).
- Find the sum of the first 12 terms (S_{12}).
Step‑by‑Step Solution
-
Common difference
[ \begin{aligned} T_{12} - T_5 &= d(12-5) \ 56 - 23 &= 7d \ 33 &= 7d \ d &= \frac{33}{7} = 4.714285714\ldots \end{aligned} ]
-
First term
Use (T_5 = a + 4d):
[ 23 = a + 4 \left( \frac{33}{7}\right) ] [ 23 = a + \frac{132}{7} ] [ a = 23 - \frac{132}{7} = \frac{161}{7} - \frac{132}{7} = \frac{29}{7} \approx 4.142857 ]
So (a = \frac{29}{7}).
-
Sum of first 12 terms
[ S_{12} = \frac{12}{2},(2a + (12-1)d) ] Plug in (a = \frac{29}{7}) and (d = \frac{33}{7}):
[ \begin{aligned} S_{12} &= 6\left( 2\cdot\frac{29}{7} + 11\cdot\frac{33}{7} \right) \ &= 6\left( \frac{58}{7} + \frac{363}{7} \right) \ &= 6\left( \frac{421}{7} \right) \ &= \frac{2526}{7} = 360.857142857\ldots \end{aligned} ]
So (S_{12} = \frac{2526}{7}) (≈ 360.86).
4. Quick Revision Summary
| Topic | Key Formula / Idea |
|---|---|
| (n^{th}) term | (T_n = a + (n-1)d) |
| Sum of first (n) terms | (S_n = \frac{n}{2},(2a + (n-1)d)) or (S_n = \frac{n}{2},(T_1+T_n)) |
| Finding (d) | (d = \frac{T_m - T_k}{m-k}) |
| Finding (a) | (a = T_k - (k-1)d) |
| Number of terms between | (\frac{y-x}{d} - 1) |
| Arithmetic Mean | (AM = \frac{a+l}{2}) (middle term) |
| Tips | • Always check if (n) is odd/even when looking for a middle term. • For sums, using the pair‑sum method (first + last, second + second‑last) often simplifies calculation. • Keep fractions in simplest form to avoid errors. |
Practice Problems (for you)
- The 3rd term of an AP is 15 and the 8th term is 32. Find (a,,d,) and (S_8).
- Find the common difference if the sum of the first 10 terms of an AP is 550 and the first term is 5.
- The 4th and 9th terms of