Find the $25$th term of the A.P. $7, 10, 13, \ldots$

Find the 25th term of the A.P. 7, 10, 13, …

Exercise: Arithmetic Progression · Q9

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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, \ \dots

The pattern is simple: you start at $a$ , then keep adding $d$ to get the next term.

Note

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 $a + 0 \cdot d$ ).

The second term is $a + d$ (that's $a + 1 \cdot d$ ).

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:

T_n = a + (n-1)d

T_n = a + (n-1)d

Example: For the AP 3, 5, 7, 9, ... we have $a = 3$ , $d = 2$ .

The 10th term: $T_{10} = 3 + (10-1) \cdot 2 = 3 + 18 = 21$ .

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 = \frac{x + z}{2}

Check: in 3, 5, 7, we have $5 = \frac{3+7}{2} = 5$ . 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: $S_n = a + (a+d) + (a+2d) + \dots + [a+(n-1)d]$

Write it backwards: $S_n = [a+(n-1)d] + [a+(n-2)d] + \dots + a$

Add them term by term. Each pair adds to $2a + (n-1)d$ , and there are $n$ such pairs. So:

2S_n = n[2a + (n-1)d]

Therefore:

S_n = \frac{n}{2}[2a + (n-1)d]

There's another useful form. Since the last term $l = a + (n-1)d$ , we can write:

S_n = \frac{n}{2}(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, ...

$S_{10} = \frac{10}{2}[2\cdot3 + (10-1)\cdot2] = 5[6 + 18] = 5 \times 24 = 120$

Quick Reference

What you need	Formula
nth term	$T_n = a + (n-1)d$
Sum of n terms	$S_n = \frac{n}{2}[2a + (n-1)d]$
Sum using last term	$S_n = \frac{n}{2}(a + l)$
Common difference	$d = T_{n+1} - T_n$

Tip

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.

Here $a=7$ , $d=10-7=3$ . So $a_{25}=a+24d=7+24(3)=7+72=79$ . [!ANSWER] $a_{25}=79$ .

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 17 on this concept.

CBSE 2025Set ANNUAL1 markMCQ
$8, 15, 22, \ldots$
(a) 8
(b) 15
(c) 22
(d) 7
›Reveal solutionSolution

The common difference $d = 7$ .

For an A.P., $d = a_2 - a_1$ (equivalently $a_3-a_2$ , etc.).

Here $a_1 = 8$ , $a_2 = 15$ : $d = 15 - 8 = 7$ . Checking: $a_3 - a_2 = 22 - 15 = 7$ , confirming a consistent common difference.

✓Final answer
$The correct option is (d) 7.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. The arithmetic mean of two numbers 12 and 15 is$
(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 $\dfrac{a+b}{2}$ .

$\dfrac{12+15}{2} = \dfrac{27}{2} = 13.5$ .

✓Final answer
$The correct option is (b) 13.5.$
CBSE 2025Set ANNUAL1 markMCQ
$5, 13, 21, 29, \ldots$
(a) 113
(b) 115
(c) 116
(d) 117
›Reveal solutionSolution

The 15th term is $117$ .

For the A.P. $5, 13, 21, 29, \ldots$ : first term $a=5$ , common difference $d=13-5=8$ .

The $n$ -th term formula is $a_n = a + (n-1)d$ . For $n=15$ : $a_{15} = 5 + (15-1)\times8 = 5 + 14\times8 = 5 + 112 = 117$ .

✓Final answer
$The correct option is (d) 117.$
CBSE 2025Set ANNUAL1 markMCQ
$9, 14, 19, 24, 29, \ldots$
(a) 80th
(b) 78th
(c) 75th
(d) 60th
›Reveal solutionSolution

Using $a_n = a + (n-1)d$ with $a=9$ , $d=5$ , and $a_n=379$ gives $n=75$ .

The A.P. is $9, 14, 19, 24, 29, \ldots$ with first term $a=9$ and common difference $d=14-9=5$ .

$a_n = a + (n-1)d$

$379 = 9 + (n-1)(5)$

$370 = 5(n-1)$

$n - 1 = 74$

$n = 75$

✓Final answer
$(c) 75th$
CBSE 2025Set ANNUAL1 markMCQ
$Q. What will be the sum of all natural numbers between 100 and 1000 which are divisible by 5?$
(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 $995$ :

$105, 110, 115, \ldots, 995$

Number of terms:

$n = \frac{995-105}{5} + 1 = \frac{890}{5}+1 = 178+1 = 179$

Sum of an A.P.:

$S_n = \frac{n}{2}(a + l) = \frac{179}{2}(105+995) = \frac{179}{2}(1100) = 179 \times 550 = 98450$

✓Final answer
$(b) 98450$
CBSE 2024Set ANNUAL1 markMCQ
$Q. Find the sum of 17 terms of the A. P. 5, 9, 13, 17, ........:$
(a) 629
(b) 529
(c) 615
(d) 0
›Reveal solutionSolution

$S_{17}=\frac{17}{2}(10+64)=629$ .

AP: $5,9,13,17,\dots$ has $a=5$ , $d=4$ .

$S_n = \dfrac n2[2a+(n-1)d]$

$S_{17} = \dfrac{17}{2}[2(5)+16(4)] = \dfrac{17}{2}[10+64] = \dfrac{17}{2}\times74 = 17\times37 = 629$ .

✓Final answer
$Sum of 17 terms = 629 — option (a).$
CBSE 2023Set ANNUAL1 markMCQ
$Q. The 10th term of a sequence whose 7th and 12th terms are 34 and 64 respectively is$
(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:

$a_7=34, \qquad a_{12}=64$

Subtracting: $a_{12}-a_7 = 5d = 64-34=30 \implies d=6$ .

The 10th term is 3 steps after the 7th term:

$a_{10} = a_7+3d = 34+3(6) = 34+18=52$

✓Final answer
$(b) 52.$
CBSE 2023Set ANNUAL1 markMCQ
$Q. The sum of odd integers from 1 to 100 is:$
(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 $n^2$ .

The odd integers between 1 and 100 are $1, 3, 5, \dots, 99$ — an AP with first term $a=1$ , common difference $d=2$ . The last term is 99, so the number of terms is:

$n = \frac{99-1}{2}+1 = 50$

Sum of first $n$ odd numbers is always $n^2$ (or use $S_n = \dfrac{n}{2}(a+l) = \dfrac{50}{2}(1+99) = 25\times100$ ):

$S = 50^2 = 2500$

✓Final answer
$(c) 2500.$
CBSE 2023Set ANNUAL1 markMCQ
$n$
(a) $\frac{n}{2}(a+b)$
(b) $\frac{n+1}{2}(a+b)$
(c) $n + 1$
(d) $\frac{n+1}{2}$
›Reveal solutionSolution

Sum of $n$ AMs $=\frac{n}{2}(a+b)$ ; 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 $\frac{a+b}{2}$ (NCERT Class 11 Sequences and Series). Hence their sum is $n\cdot\frac{a+b}{2}=\frac{n}{2}(a+b)$ .

✓Final answer
$\frac{n}{2}(a+b)$
CBSE 2022Set ANNUAL1 mark
$Q. If nth term of an A.P. is 5n + 1, then its common difference is ............ .$
›Reveal solutionSolution

For $a_n = 5n + 1$ , consecutive terms differ by the coefficient of $n$ , which is $5$ .

Given $a_n = 5n + 1$ .

$d = a_{n+1} - a_n = [5(n+1) + 1] - [5n + 1] = 5n + 5 + 1 - 5n - 1 = 5$

✓Final answer
$5$
CBSE 2022Set ANNUAL1 mark
$Q. If 5th term of an A.P. is 10 and 10th term is 5, then its 15th term is ............ .$
›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 $n$ th term is $a_n = a + (n-1)d$ .

5th term: $a + 4d = 10$

10th term: $a + 9d = 5$

Subtracting: $(a+9d) - (a+4d) = 5 - 10 \Rightarrow 5d = -5 \Rightarrow d = -1$

From the first equation: $a = 10 - 4d = 10 - 4(-1) = 14$

15th term:

$a_{15} = a + 14d = 14 + 14(-1) = 14 - 14 = 0$

✓Final answer
$0$
CBSE 2022Set ANNUAL1 markMCQ
$\_\_\_\_$
(a) $5\sqrt{2}$
(b) 8
(c) $7\sqrt{2}$
›Reveal solutionSolution

Rewriting each term as a multiple of $\sqrt2$ reveals an A.P., whose 5th term is $5\sqrt2$ .

$\sqrt2 = \sqrt2$

$\sqrt8 = \sqrt{4\times2} = 2\sqrt2$

$\sqrt{18} = \sqrt{9\times2} = 3\sqrt2$

So the series is $\sqrt2, 2\sqrt2, 3\sqrt2, \ldots$ , an A.P. with first term $a=\sqrt2$ and common difference $d=\sqrt2$ .

The 5th term $= a+4d = \sqrt2+4\sqrt2 = 5\sqrt2$ .

✓Final answer
$5\sqrt2$

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Q.Find the 252525th term of the A.P. 7,10,13,…7, 10, 13, \ldots7,10,13,…

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Q.Find the $25$ th term of the A.P. $7, 10, 13, \ldots$

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