The sum of terms equidistant from the beginning and end in an A.P. is equal to ............ .

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Q. The sum of terms equidistant from the beginning and end in an A.P. is equal to ............ .

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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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In any arithmetic progression, terms that are equidistant from the beginning and end always add up to the same value: the sum of the first and last terms. This sum equals $a + l$ (or $2a + (n-1)d$ ).

The beauty of an arithmetic progression lies in its symmetry. When you pick any two terms that are the same distance from opposite ends, they balance each other perfectly around the middle of the sequence.

Think of it this way: as you move forward from the first term, each step adds the common difference $d$ . As you move backward from the last term, each step subtracts the same $d$ . These changes cancel out when you add the two terms together.

Let me show you why this works with the general form of an A.P.: $a, a+d, a+2d, \ldots, a+(n-1)d$ , where $a$ is the first term, $d$ is the common difference, $n$ is the number of terms, and $l = a + (n-1)d$ is the last term.

Pick the $r$ -th term from the beginning.

The $r$ -th term is $T_r = a + (r-1)d$ .
Pick the $r$ -th term from the end.

Counting backward, the $r$ -th term from the end is the $(n-r+1)$ -th term from the beginning.

So $T_{n-r+1} = a + (n-r+1-1)d = a + (n-r)d$ .
Add these two equidistant terms.

T_r + T_{n-r+1} = [a + (r-1)d] + [a + (n-r)d]

= 2a + (r-1)d + (n-r)d

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

= 2a + (n-1)d

Recognize what this equals. …

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

CA Foundation 2026Set may-20261 markMCQ
$4n^2 + 3n$
›Reveal solutionSolution

Use $a_n=S_n-S_{n-1}$ : $a_{10}=S_{10}-S_9=430-351=79$ .

Step 1 — Recall the relation

The $n$ -th term of any sequence equals the difference of consecutive partial sums:
$a_n=S_n-S_{n-1}$
Step 2 — Evaluate $S_{10}$ and $S_9$
$S_{10}=4(10)^2+3(10)=400+30=430$ $S_9=4(9)^2+3(9)=324+27=351$
Step 3 — Subtract

$a_{10}=430-351=79$ …
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CA Foundation 2026Set may-20261 markMCQ
$4^{th}$
›Reveal solutionSolution

Terms equidistant from a middle term average to it: $a_4+a_8=2a_6$ , so $a_6=120/2=60$ .

Step 1 — Write the two terms
$a_4=a+3d,\qquad a_8=a+7d$
Step 2 — Add them
$a_4+a_8=(a+3d)+(a+7d)=2a+10d=2(a+5d)$
But $a+5d=a_6$ , so:
$a_4+a_8=2a_6$
Step 3 — Solve

$2a_6=120\ \Rightarrow\ a_6=60$ …
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CA Foundation 2025Set jan-20251 markMCQ
$Q. The sum of the 4th and 8th term of an AP is 10. Then the sum of first eleven terms of the series is (A) 33 (B) 22 (C) 44 (D) 55$
›Reveal solutionSolution

$t_4 + t_8 = 2a + 10d = 10$ , and $S_{11} = \frac{11}{2}(2a+10d) = \frac{11}{2}(10) = 55$ .

Step 1 — Express the two terms
$t_4 = a + 3d, \qquad t_8 = a + 7d$ $t_4 + t_8 = 2a + 10d = 10$
Step 2 — Write the sum of 11 terms
$S_n = \frac{n}{2}\big(2a + (n-1)d\big)$
Step 3 — Substitute the known value
$S_{11} = \frac{11}{2}(10) = 11 \times 5 = 55$
Why the other options are wrong: 33, 22 and 44 come from using a wrong $n$ or mis-forming $2a+(n-1)d$ . …
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CA Foundation 2025Set jan-20251 markMCQ
$8, 5, 2, -1, -4, \ldots\ldots$
›Reveal solutionSolution

$a = 8$ , $d = -3$ , so $t_9 = 8 + (9-1)(-3) = -16$ .

Step 1 — Identify $a$ and $d$
$a = 8, \qquad d = 5 - 8 = -3$
Step 2 — Apply the nth-term formula
$t_n = a + (n-1)d$ $t_9 = 8 + (9-1)(-3) = 8 + 8(-3)$
Step 3 — Evaluate
$t_9 = 8 - 24 = -16$
Why the other options are wrong: (A) $-10$ uses $d=-2$ ; (B) $-24$ forgets the $+8$ ; (D) $-4$ stops at the 5th term. …
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CA Foundation 2025Set jan-20251 markMCQ
$1 + 2 + 3 + \ldots\ldots$
›Reveal solutionSolution

$\frac{n(n+1)}{2} = 55 \Rightarrow n(n+1) = 110 \Rightarrow n = 10$ .

Step 1 — Use the sum of first $n$ natural numbers
$S_n = \frac{n(n+1)}{2}$
Step 2 — Set equal to 55 and simplify
$\frac{n(n+1)}{2} = 55 \;\Rightarrow\; n(n+1) = 110$
Step 3 — Solve the quadratic
$n^2 + n - 110 = 0 \;\Rightarrow\; (n-10)(n+11) = 0$
Taking the positive root, $n = 10$ . …
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CA Foundation 2025Set may-20251 markMCQ
$n^{th}$
›Reveal solutionSolution

$S_n=\frac{n}{2}(a_1+a_n)=\frac{n}{2}(6+5n+1)=\frac{n(5n+7)}{2}$ .

Step 1 — Find the first term

Given $a_n=5n+1$ , put $n=1$ : $a_1=5(1)+1=6$ .

Step 2 — Apply the A.P. sum formula
$S_n=\dfrac{n}{2}\,(a_1+a_n)$
Step 3 — Substitute and simplify
$S_n=\dfrac{n}{2}\big(6+(5n+1)\big)=\dfrac{n}{2}(5n+7)=\dfrac{n(7+5n)}{2}$
Why the other options are wrong: (D) $\frac{n(7+4n)}{2}$ uses the wrong common-difference term (coefficient 4 instead of 5); (A) and (B) are unrelated single fractions. …
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CA Foundation 2025Set may-20251 markMCQ
$Q. Insert 4 numbers between 2 and 22 such that the resulting sequence is an Arithmetic Progression (A.P.). (A) 4, 8, 12, 16 (B) 5, 9, 13, 17 (C) 4, 10, 15, 19 (D) 6, 10, 14, 18$
›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 $a_1=2$ and $a_6=22$ .

Step 2 — Find the common difference
$a_6=a_1+5d \Rightarrow 22=2+5d \Rightarrow d=\dfrac{20}{5}=4$
Step 3 — Build the sequence
$2,\;6,\;10,\;14,\;18,\;22$
The four inserted (arithmetic mean) numbers are $6,10,14,18$ . …
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CA Foundation 2025Set sep-20251 markMCQ
$\dfrac{1}{3}, \dfrac{1 - 3b}{3}, \dfrac{1 - 6b}{3}, \ldots$
›Reveal solutionSolution

$d = t_2 - t_1 = \dfrac{1-3b}{3} - \dfrac{1}{3} = -b$ .

Step 1 — Recall the definition
$d = t_{k+1} - t_k \quad(\text{same for every consecutive pair})$
Step 2 — Subtract consecutive terms
$d = \frac{1-3b}{3} - \frac{1}{3} = \frac{(1-3b) - 1}{3} = \frac{-3b}{3} = -b$
Step 3 — Confirm with the next pair
$\frac{1-6b}{3} - \frac{1-3b}{3} = \frac{-3b}{3} = -b \ \checkmark$
The difference is constant, confirming it is a valid AP with $d = -b$ . …
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CA Foundation 2025Set sep-20251 markMCQ
$Q. The sum of all natural numbers between 200 and 600 those are divisible by 13 is __________. (A) 12493 (B) 14493 (C) 16493 (D) 18493$
›Reveal solutionSolution

Multiples of 13 in range: $208$ to $598$ , 31 terms; $S = \frac{31}{2}(208+598) = 12{,}493$ .

Step 1 — Find the first and last multiples of 13

$13 \times 16 = 208$ (first multiple above 200) and $13 \times 46 = 598$ (last multiple below 600).

Step 2 — Count the terms
$n = 46 - 16 + 1 = 31$
Step 3 — Sum the arithmetic series
$S_n = \frac{n}{2}(a + l)$ $S_{31} = \frac{31}{2}(208 + 598) = \frac{31}{2}\times 806 = 31 \times 403 = 12{,}493$
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. …
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Q.The sum of terms equidistant from the beginning and end in an A.P. is equal to ............ .

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