The 15th term of the A.P. $5, 13, 21, 29, \ldots$ is (a) 113 (b) 115 (c) 116 (d) 117

The 15th term of the A.P. 5, 13, 21, 29, … is (a) 113 (b) 115 (c) 116 (d) 117

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5, 13, 21, 29, \ldots

(a) 113

(b) 115

(d) 117

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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	$d = T_{n+1} - T_n$

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The 15th term is $117$ .

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

…

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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.).

…

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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 $\dfrac{a+b}{2}$ .

…

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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 $d=13-5=8$ .

…

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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 $d=14-9=5$ .

$a_n = a + (n-1)d$ …

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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: …

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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 $a=5$ , $d=4$ .

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

…

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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:

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

…

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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:

$n = \frac{99-1}{2}+1 = 50$ …

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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 a …

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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 $a_n = 5n + 1$ .

…

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

…

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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 $5\sqrt2$ .

$\sqrt2 = \sqrt2$

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

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

…

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Q.The 15th term of the A.P. 5,13,21,29,…5, 13, 21, 29, \ldots5,13,21,29,… is

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Q.The 15th term of the A.P. $5, 13, 21, 29, \ldots$ is

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