Q.The first term of an A.P. is $a$, and the sum of the first $p$ terms is zero, show that the sum of its next $q$ terms is $\dfrac{-a(p+q)q}{p-1}$. [Hint: Required sum $= S_{p+q} - S_p$]

Showing the 12 most recent of 13 on this concept.

1. CA Foundation 2026Set may-20261 markMCQ
   Q.The sum of the first n terms of an arithmetic progression (A.P.) is $4n^2 + 3n$. The $10^{th}$ term of the A.P. is ______. (A) 77 (B) 83 (C) 81 (D) 79

   ›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

   $a_{10}=430-351=79$

   Tip

   Alternatively, $S_n=4n^2+3n$ has the form of an AP sum with common difference $d=8$ and first term $a_1=S_1=7$, so $a_{10}=7+9(8)=79$ — same answer.

   Watch out

   Don't just plug $n=10$ into $S_n$ — that gives the sum of ten terms (430), not the tenth term. You must subtract $S_9$.

   ✓Final answer

   (D) 79

2. CA Foundation 2026Set may-20261 markMCQ
   Q.If the sum of $4^{th}$ and $8^{th}$ term of an arithmetic progression (A.P.) is 120, then the $6^{th}$ term of the A.P. is ______. (A) 10 (B) 70 (C) 60 (D) 100

   ›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

   $2a_6=120\ \Rightarrow\ a_6=60$

   Tip

   In an AP, the sum of two terms equidistant from a term equals twice that term (here $4$ and $8$ are symmetric about $6$). This shortcut avoids solving for $a$ and $d$ separately.

   Watch out

   Don't try to find $a$ and $d$ individually — one equation cannot pin down both, and you don't need to. The symmetry gives $a_6$ directly.

   ✓Final answer

   (C) 60

3. COMEDK 2025Set 2025-E1 markMCQ
   Q.Given that n number of arithmetic means are inserted between two pairs of numbers $a, 2 b$ and $2 a, b$; where $a, b \in R$. If the $m^{\text {th }}$ means in the two cases are the same, then the ratio $a: b$ is equal to (A) $m:(n-m+1)$ (B) $n:(n-m+1)$ (C) $(n-m+1): m$ (D) $(n-m+1): n$

   ›Reveal solutionSolution

   The key idea is to write the $m^{\text{th}}$ arithmetic mean in each sequence using the formula for the $k^{\text{th}}$ term of an AP, then equate them. The ratio simplifies to $a:b = m:(n-m+1)$, which corresponds to option (A).

   ---

   We are inserting $n$ arithmetic means between two numbers. That means we are creating an arithmetic progression (AP) that starts at the first number, ends at the second number, and has $n$ terms in between. The $m^{\text{th}}$ mean is the $m^{\text{th}}$ term after the first number, so it is the $(m+1)^{\text{th}}$ term of the full AP.

   Why this approach works:

   If we know the first term $A$ and the last term $B$ of an AP with $n$ means inserted, the common difference $d$ is $\frac{B-A}{n+1}$. Then any mean (the $k^{\text{th}}$ one) is $A + kd$. Equating the $m^{\text{th}}$ means from two different sequences gives a direct equation in $a$ and $b$, from which the ratio follows.

   ---

   Step-by-step solution:

   1. First case: numbers $a$ and $2b$, with $n$ means inserted. The AP has first term $a$, last term $2b$, and total terms $n+2$. Common difference:

   $d_1 = \frac{2b - a}{n+1}$

   The $m^{\text{th}}$ mean is the $(m+1)^{\text{th}}$ term:

   $M_1 = a + m d_1 = a + m \cdot \frac{2b - a}{n+1}$

   2. Second case: numbers $2a$ and $b$, with $n$ means inserted. First term $2a$, last term $b$, common difference:

   $d_2 = \frac{b - 2a}{n+1}$

   The $m^{\text{th}}$ mean:

   $M_2 = 2a + m d_2 = 2a + m \cdot \frac{b - 2a}{n+1}$

   3. Set them equal (given that the $m^{\text{th}}$ means are the same):

   $a + \frac{m(2b - a)}{n+1} = 2a + \frac{m(b - 2a)}{n+1}$

   4. Simplify: Multiply both sides by $(n+1)$:

   $a(n+1) + m(2b - a) = 2a(n+1) + m(b - 2a)$

   Expand:

   $a n + a + 2mb - m a = 2a n + 2a + m b - 2m a$

   5. Collect like terms. Bring everything to one side:

   $(a n + a - 2a n - 2a) + (2mb - m b) + (-m a + 2m a) = 0$

   Simplify each group:

   • $a$ terms: $a n + a - 2a n - 2a = -a n - a$
   • $b$ terms: $2mb - m b = m b$
   • $a$ terms from $m$: $-m a + 2m a = m a$

   So we have:

   $-a n - a + m b + m a = 0$

   6. Factor and rearrange:

   $-a(n+1) + m(b + a) = 0$

   $m(a + b) = a(n+1)$

   $\frac{a+b}{a} = \frac{n+1}{m}$

   But we want $a:b$, so write $b$ in terms of $a$:

   $m(a+b) = a(n+1) \quad\Rightarrow\quad m b = a(n+1 - m)$

   Hence:

   $\frac{a}{b} = \frac{m}{n+1 - m}$

   7. Interpret the denominator: $n+1 - m$ is exactly $(n - m + 1)$. So:

   $a:b = m : (n - m + 1)$

   Tip

   Notice that $n+1$ is the number of gaps between the $n+2$ terms. The $m^{\text{th}}$ mean is $m$ steps from the start, leaving $n+1-m$ steps to the end — that’s why the ratio appears so neatly.

   Watch out

   A common mistake is to forget that the $m^{\text{th}}$ mean is the $(m+1)^{\text{th}}$ term, not the $m^{\text{th}}$ term of the full AP. Using $m$ instead of $m+1$ in the term index would give a different (wrong) ratio.

   ---

   ✓Final answer

   The correct option is (A).

   ANSWER: A

4. 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

   $S_n = \frac{n}{2}\big(2a + (n-1)d\big)$

   For $n = 11$: $S_{11} = \frac{11}{2}(2a + 10d)$.

   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 $2a+(n-1)d$.

   Watch out

   $2a + (11-1)d = 2a + 10d$ — exactly the quantity you were given. Don't waste time solving for $a$ and $d$ separately; they aren't individually determined.

   Tip

   $S_{11} = 11 \times t_6$ (the middle term), and $t_6 = \frac{t_4 + t_8}{2} = 5$, so $S_{11} = 55$ instantly.

   ✓Final answer

   (D) 55

5. CA Foundation 2025Set jan-20251 markMCQ
   Q.Find the 9th term of the A.P. $8, 5, 2, -1, -4, \ldots\ldots$ (A) $-10$ (B) $-24$ (C) $-16$ (D) $-4$

   ›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

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

   For $n = 9$:

   $t_9 = 8 + (9-1)(-3) = 8 + 8(-3)$

   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.

   Watch out

   Use $(n-1)$ steps of $d$, not $n$. For the 9th term you add $d$ eight times, giving $8 \times (-3) = -24$.

   Tip

   With a negative $d$, list a couple more terms mentally ($-4, -7, -10, -13, -16$) as a quick check on $t_9$.

   ✓Final answer

   (C) $-16$

6. CA Foundation 2025Set jan-20251 markMCQ
   Q.The sum of series $1 + 2 + 3 + \ldots\ldots$ is 55. The number of terms is : (A) 40 (B) 30 (C) 20 (D) 10

   ›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, $n = 10$.

   Why the other options are wrong: 20, 30 and 40 give sums of 210, 465 and 820 respectively — far above 55.

   Watch out

   Reject the negative root $n = -11$: a count of terms must be a positive whole number.

   Tip

   Recognise 55 as a triangular number ($T_{10} = 55$) to jump straight to $n = 10$.

   ✓Final answer

   (D) 10

7. CA Foundation 2025Set may-20251 markMCQ
   Q.Find the sum of n terms of the A.P., whose $n^{th}$ term is $5n + 1$. (A) $\frac{n}{2}$ (B) $\frac{2n}{7}$ (C) $\frac{n(7+5n)}{2}$ (D) $\frac{n(7+4n)}{2}$

   ›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.

   Watch out

   Do not confuse the $n^{th}$ term with the sum — first extract $a_1$ (and, if needed, the common difference $d=5$) before summing.

   Tip

   When $a_n$ is linear in $n$, $S_n=\frac{n}{2}(a_1+a_n)$ is the fastest route — no need to find $d$ explicitly.

   ✓Final answer

   (C) $\dfrac{n(7+5n)}{2}$

8. 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 $6,10,14,18$.

   Why the other options are wrong: (A) 4,8,12,16 uses $d=4$ but starts from the wrong first inserted value; (B) and (C) do not form a constant-difference sequence ending at 22.

   Watch out

   There are 5 gaps (not 4) between the 6 terms — divide the total span 20 by 5, not by 4.

   Tip

   Inserting $k$ arithmetic means between $a$ and $b$: $d=(b-a)/(k+1)$.

   ✓Final answer

   (D) 6, 10, 14, 18

9. CA Foundation 2025Set sep-20251 markMCQ
   Q.The common difference of the arithmetic progression $\dfrac{1}{3}, \dfrac{1 - 3b}{3}, \dfrac{1 - 6b}{3}, \ldots$ is __________. (A) $-b$ (B) $b$ (C) $-3b$ (D) $3b$

   ›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 $d = -b$.

   Why the other options are wrong: (B) $b$ has the wrong sign; (C) $-3b$ and (D) $3b$ forget to divide the numerator difference by 3.

   Watch out

   Keep the common denominator 3 through the subtraction — dropping it gives $-3b$ (option C), a classic trap.

   Tip

   For an AP written with a common denominator, just subtract numerators, then divide once.

   ✓Final answer

   (A) $-b$

10. 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.

    Watch out

    "Between 200 and 600" excludes both endpoints — start at 208, not 195, and stop at 598, not 611.

    Tip

    Convert the divisibility count into an AP: index the multiples ($13\times16$ to $13\times46$) and the term count is just the difference of the indices plus one.

    ✓Final answer

    (A) 12493

11. COMEDK 2023Set 2023-M1 markMCQ
    Q.If the sum of 12th and 22nd terms of an AP is 100, then the sum of the first 33 terms of an $\mathrm{AP}$ is (A) 1700 (B) 1650 (C) 3300 (D) 3500

    ›Reveal solutionSolution

    $T_{12}+T_{22}=100$ gives $a+16d=50$; the 33-term sum is $33(a+16d)=1650$.

    Let the AP have first term $a$ and common difference $d$.

    $T_{12}+T_{22}=(a+11d)+(a+21d)=2a+32d=100\;\Rightarrow\;a+16d=50.$

    The sum of the first 33 terms:

    $S_{33}=\frac{33}{2}\bigl(2a+(33-1)d\bigr)=\frac{33}{2}(2a+32d)=33\,(a+16d)=33\cdot50=1650.$

    (Note $a+16d=T_{17}$, the middle term of the 33 terms.)

    ✓Final answer

    The correct option is (B) — 1650

12. COMEDK 2022Set 20221 markMCQ
    Q.The first and fifth terms of an A.P. are $-14$ and 2 respectively and the sum of its n terms is 40. The value of n is (A) 8 (B) 12 (C) 10 (D) 13

    ›Reveal solutionSolution

    Check: S₁₀ = 2(100) − 160 = 40 ✓.

    Concept: AP — nth term and sum formulas.

    a = −14, a₅ = a + 4d = 2 → −14 + 4d = 2 → 4d = 16 → d = 4.

    Sₙ = (n/2)[2a + (n−1)d] = (n/2)[−28 + 4(n−1)] = (n/2)(4n − 32) = 2n² − 16n.

    Set Sₙ = 40:

    2n² − 16n − 40 = 0 → n² − 8n − 20 = 0 → (n − 10)(n + 2) = 0 → n = 10 (n = −2 rejected).

    Check: S₁₀ = 2(100) − 160 = 40 ✓.

    ✓Final answer

    The correct option is (C) — 10

    ANSWER: C