Sequences & Series

Difficulty Level: 1 (Recognition — MCQ, Matching, True/False with justification) Time Limit: 20 minutes Total Marks: 30

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Section A — Multiple Choice Questions (1 mark each)

Choose the correct option.

Q1. The $n$th term of an AP with first term $a$ and common difference $d$ is: (a) $a + nd$ (b) $a + (n-1)d$ (c) $an + d$ (d) $a - (n-1)d$

Q2. The sum of the first $n$ natural numbers $\sum_{k=1}^{n} k$ equals: (a) $\dfrac{n(n+1)}{2}$ (b) $\dfrac{n(n-1)}{2}$ (c) $n^2$ (d) $\dfrac{n(2n+1)}{6}$

Q3. An infinite GP $a + ar + ar^2 + \cdots$ converges if and only if: (a) $r > 1$ (b) $|r| < 1$ (c) $r < 0$ (d) $|r| \ge 1$

Q4. The sum of an infinite convergent GP with first term $a$ and ratio $r$ is: (a) $\dfrac{a}{1-r}$ (b) $\dfrac{a}{r-1}$ (c) $\dfrac{a(1-r)}{1}$ (d) $\dfrac{1}{1-r}$

Q5. If $a, b, c$ are in HP, then $b$ (the Harmonic Mean of $a$ and $c$) equals: (a) $\dfrac{a+c}{2}$ (b) $\sqrt{ac}$ (c) $\dfrac{2ac}{a+c}$ (d) $\dfrac{ac}{a+c}$

Q6. For two positive numbers, the correct inequality is: (a) $AM \le GM \le HM$ (b) $HM \le GM \le AM$ (c) $GM \le AM \le HM$ (d) $AM \le HM \le GM$

Q7. The value of $\sum_{k=1}^{n} k^2$ is: (a) $\dfrac{n(n+1)}{2}$ (b) $\left(\dfrac{n(n+1)}{2}\right)^2$ (c) $\dfrac{n(n+1)(2n+1)}{6}$ (d) $\dfrac{n(n+1)(n+2)}{6}$

Q8. The number of terms in the binomial expansion of $(x+y)^n$ is: (a) $n$ (b) $n-1$ (c) $n+1$ (d) $2n$

Q9. The general term $T_{r+1}$ in the expansion of $(x+y)^n$ is: (a) $\binom{n}{r} x^{n-r} y^r$ (b) $\binom{n}{r} x^r y^{n-r}$ (c) $\binom{n}{r+1} x^{n-r} y^r$ (d) $\binom{n}{r} x^n y^r$

Q10. The relation between AM, GM, HM of two positive numbers is: (a) $GM^2 = AM \cdot HM$ (b) $AM^2 = GM \cdot HM$ (c) $HM^2 = AM \cdot GM$ (d) $AM = GM = HM$ always

Q11. A telescoping sum $\sum_{k=1}^{n}\left(\dfrac{1}{k} - \dfrac{1}{k+1}\right)$ equals: (a) $1 - \dfrac{1}{n+1}$ (b) $\dfrac{1}{n}$ (c) $1 + \dfrac{1}{n}$ (d) $\dfrac{1}{n+1}$

Q12. In mathematical induction, after verifying the base case, we must: (a) prove the statement for all $n$ directly (b) assume $P(k)$ true and prove $P(k+1)$ (c) assume $P(k+1)$ and prove $P(k)$ (d) check only $n=1$ and $n=2$

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Section B — Matching (1 mark each; total 5 marks)

Q13. Match each expression in Column I with its value/form in Column II.

Column I	Column II
(i) $\sum_{k=1}^{n} 1$	(P) $\dfrac{n(n+1)(2n+1)}{6}$
(ii) $\sum_{k=1}^{n} k^3$	(Q) $n$
(iii) $\sum_{k=1}^{n} k^2$	(R) $\dfrac{a(r^n-1)}{r-1}$
(iv) Sum of $n$ terms of GP $(r\ne1)$	(S) $\left(\dfrac{n(n+1)}{2}\right)^2$
(v) GM of $a$ and $b$	(T) $\sqrt{ab}$

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Section C — True / False WITH Justification (2 marks each; total 13 marks)

State True or False and give a one-line reason. 1 mark for verdict, 1 mark for justification.

Q14. The sequence $2, 4, 8, 16, \ldots$ is an arithmetic progression.

Q15. The infinite series $1 + 2 + 4 + 8 + \cdots$ has a finite sum.

Q16. In Pascal's triangle, the entry in row $n$, position $r$ (both from 0) equals $\binom{n}{r}$, and $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$.

Q17. For a rational index, $(1+x)^{1/2} \approx 1 + \dfrac{x}{2}$ is a valid approximation for small $x$.

Q18. For any two distinct positive reals, $AM = GM$.

Q19. The middle term of $(x+y)^6$ is the 4th term.

Q20. Strong induction assumes the statement holds for all values from the base up to $k$ to prove it for $k+1$.

[
  {"claim":"HM of a and c equals 2ac/(a+c)","code":"a,c=symbols('a c',positive=True); HM=2/(1/a+1/c); result = simplify(HM-2*a*c/(a+c))==0"},
  {"claim":"GM^2 = AM*HM for two positives","code":"a,b=symbols('a b',positive=True); AM=(a+b)/2; HM=2*a*b/(a+b); GM2=a*b; result = simplify(GM2-AM*HM)==0"},
  {"claim":"Telescoping sum equals 1-1/(n+1)","code":"k,n=symbols('k n',positive=True,integer=True); S=summation(1/k-1/(k+1),(k,1,n)); result = simplify(S-(1-1/(n+1)))==0"},
  {"claim":"Sum of k^2 equals n(n+1)(2n+1)/6","code":"k,n=symbols('k n',positive=True,integer=True); S=summation(k**2,(k,1,n)); result = simplify(S-n*(n+1)*(2*n+1)/6)==0"}
]