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CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Model Questions - 1 Marks - Part 1

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SECTION A — Multiple Choice Questions [1 Mark Each]

• 1 Mark Q1. If the $p^{\text{th}}$ term of an AP is $q$ and its $q^{\text{th}}$ term is $p$, then its $n^{\text{th}}$ term is:
  (a).$p + q - n$

  (b).$p + q + n$

  (c).$p - q + n$

  (d).$p - q - n$
  Solution:
  Let first term be $a$ and common difference be $d$.
  $a_p = a + (p-1)d = q$ and $a_q = a + (q-1)d = p$.
  Subtracting the two equations gives $(p-q)d = q - p \implies d = -1$.
  Substituting $d = -1$ into $a_p$, $a = q + p - 1$.
  $a_n = a + (n-1)d = (q + p - 1) + (n-1)(-1) = p + q - n$.
  Answer: (a) $p + q - n$
• 1 Mark Q2. How to evaluate the expression $\frac{a - b}{b - c}$ when $a, b,$ and $c$ are in Arithmetic Progression (AP).
  (a).$1$

  (b).$\frac{a}{c}$

  (c).$\frac{b}{c}$

  (d).$2$
  Solution:
  Since $a, b, c$ are in AP, the common difference is constant: $b - a = c - b \implies a - b = b - c$.
  Therefore, $\frac{a - b}{b - c} = 1$.
  Answer: (a) $1$
• 1 Mark Q3. The sum of the first $n$ terms of an AP is given by $S_n = 3n^2 + 5n$. Which term of this AP is $164$?
  (a).$27^{\text{th}}$ term

  (b).$26^{\text{th}}$ term

  (c).$28^{\text{th}}$ term

  (d).$25^{\text{th}}$ term
  Solution:
  $a_n = S_n - S_{n-1} = (3n^2 + 5n) - [3(n-1)^2 + 5(n-1)] = 6n + 2$.
  Set $a_n = 164 \implies 6n + 2 = 164 \implies 6n = 162 \implies n = 27$.
  Answer: (a) $27^{\text{th}}$ term
• 1 Mark Q4. If $m$ times the $m^{\text{th}}$ term of an AP is equal to $n$ times its $n^{\text{th}}$ term ($m \neq n$), then its $(m + n)^{\text{th}}$ term is:
  (a).$0$

  (b).$m + n$

  (c).$m - n$

  (d).$1$
  Solution:
  Given $m \cdot a_m = n \cdot a_n \implies m[a + (m-1)d] = n[a + (n-1)d]$.
  Simplifying yields $a(m-n) + d(m^2 - n^2) = 0 \implies (m-n)[a + (m+n-1)d] = 0$.
  Since $m \neq n$, $a + (m+n-1)d = 0$, which is the $(m+n)^{\text{th}}$ term.
  Answer: (a) $0$
• 1 Mark Q5. If the sum of first $n$ terms of an AP is $S_n = 2n^2 + 3n$, then its common difference $d$ is:
  (a).$4$

  (b).$2$

  (c).$3$

  (d).$5$
  Solution:
  $a_1 = S_1 = 2(1)^2 + 3(1) = 5$.
  $S_2 = 2(2)^2 + 3(2) = 14 \implies a_2 = S_2 - S_1 = 14 - 5 = 9$.
  Common difference $d = a_2 - a_1 = 9 - 5 = 4$.
  Answer: (a) $4$
• 1 Mark Q6. If $\frac{1}{x+2}, \frac{1}{x+3}, \frac{1}{x+5}$ are in AP, then $x$ is equal to:
  (a).$1$

  (b).$2$

  (c).$3$

  (d).$5$
  Solution:
  Using the property of AP terms: $2b = a + c \implies \frac{2}{x+3} = \frac{1}{x+2} + \frac{1}{x+5}$.
  $\frac{2}{x+3} = \frac{(x+5) + (x+2)}{(x+2)(x+5)} \implies 2(x^2 + 7x + 10) = (x+3)(2x+7)$.
  $2x^2 + 14x + 20 = 2x^2 + 13x + 21 \implies x = 1$.
  Answer: (a) $1$
• 1 Mark Q7. The sum of all 2-digit natural numbers which leave a remainder $1$ when divided by $4$ is:
  (a).$1210$

  (b).$1200$

  (c).$1250$

  (d).$1180$
  Solution:
  The sequence is $13, 17, 21, \dots, 97$.
  $a = 13, d = 4, l = 97$. Number of terms $n = \frac{97 - 13}{4} + 1 = 22$.
  Sum $S_{22} = \frac{22}{2}(13 + 97) = 11 \times 110 = 1210$.
  Answer: (a) $1210$
• 1 Mark Q8. If the $11^{\text{th}}$ term of an AP is $38$ and the $16^{\text{th}}$ term is $73$, then its $31^{\text{st}}$ term is:
  (a).$178$

  (b).$168$

  (c).$188$

  (d).$158$
  Solution:
  $a + 10d = 38$ and $a + 15d = 73$.
  Subtracting gives $5d = 35 \implies d = 7$, and $a = -32$.
  $a_{31} = a + 30d = -32 + 30(7) = 178$.
  Answer: (a) $178$
• 1 Mark Q9. If $S_n$ denotes the sum of first $n$ terms of an AP, then $S_{3n}$ is equal to:
  (a).$3(S_{2n} - S_n)$

  (b).$2(S_{2n} - S_n)$

  (c).$S_{2n} - S_n$

  (d).$4(S_{2n} - S_n)$
  Solution:
  $S_{2n} - S_n$ is the sum of terms from $(n+1)$ to $2n$. $S_{3n} - S_{2n}$ is the sum from $(2n+1)$ to $3n$.
  In an AP, consecutive sums of equal blocks of terms are in AP themselves, but using formulas or standard identities: $3(S_{2n} - S_n)$ equals $S_{3n}$ under specific formulations or check standard identity for AP sums. Actually, let's verify: $S_{2n} - S_n$ represents the middle block, and $S_{3n}$ relation can be derived as $3(S_{2n} - S_n)$.
  Answer: (a) $3(S_{2n} - S_n)$
• 1 Mark Q10. How many terms of the AP $24, 21, 18, \dots$ must be taken so that their sum is $78$?
  (a).$4$ or $13$

  (b).$5$ or $12$

  (c).$6$ or $11$

  (d).$4$ only
  Solution:
  $S_n = \frac{n}{2}[2a + (n-1)d] \implies 78 = \frac{n}{2}[48 + (n-1)(-3)]$.
  $156 = n(51 - 3n) \implies 3n^2 - 51n + 156 = 0 \implies n^2 - 17n + 52 = 0$.
  $(n - 4)(n - 13) = 0 \implies n = 4$ or $13$.
  Answer: (a) $4$ or $13$
• 1 Mark Q11. If the sum of the first $n$ terms of an AP is $S_n = An + Bn^2$, where $A$ and $B$ are constants, then its common difference is:
  (a).$2B$

  (b).$2A$

  (c).$A + B$

  (d).$A - B$
  Solution:
  $a_n = S_n - S_{n-1} = (An + Bn^2) - [A(n-1) + B(n-1)^2] = A + B(2n - 1)$.
  $a_{n+1} - a_n = 2B$, which is the common difference.
  Answer: (a) $2B$
• 1 Mark Q12. The sum of the first $p$ terms of an AP is $q$ and the sum of the first $q$ terms is $p$. The sum of its first $(p + q)$ terms is:
  (a).$-(p + q)$

  (b).$p + q$

  (c).$p - q$

  (d).$0$
  Solution:
  $S_{p+q} = \frac{p+q}{2}[2a + (p+q-1)d]$. Standard property yields $S_{p+q} = -(p+q)$.
  Answer: (a) $-(p + q)$
• 1 Mark Q13. If $a, b, c, d, e$ are in AP, then the value of $a - 4b + 6c - 4d + e$ is:
  (a).$0$

  (b).$1$

  (c).$2c$

  (d).$a + e$
  Solution:
  Using binomial coefficients with alternating signs for an AP terms: $a - 4(a+d) + 6(a+2d) - 4(a+3d) + (a+4d) = a(1-4+6-4+1) + d(-4+12-12+4) = 0$.
  Answer: (a) $0$
• 1 Mark Q14. The ratio of the sum of $n$ terms of two arithmetic progressions is $(7n + 1) : (4n + 27)$. The ratio of their $11^{\text{th}}$ terms is:
  (a).$4 : 3$

  (b).$3 : 4$

  (c).$2 : 3$

  (d).$5 : 6$
  Solution:
  Replace $n$ by $21$ to find the $11^{\text{th}}$ term ratio: $\frac{7(21) + 1}{4(21) + 27} = \frac{148}{111} = \frac{4}{3}$.
  Answer: (a) $4 : 3$
• 1 Mark Q15. If the $n^{\text{th}}$ term of an AP is $a_n = 2n + 1$, then the sum of its first $n$ terms is:
  (a).$n(n + 2)$

  (b).$n(n + 1)$

  (c).$n^2$

  (d).$2n^2$
  Solution:
  $a_1 = 3, a_2 = 5$. $S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}(3 + 2n + 1) = n(n + 2)$.
  Answer: (a) $n(n + 2)$
• 1 Mark Q16. If $k, 2k - 1, 2k + 1$ are three consecutive terms of an AP, then the value of $k$ is:
  (a).$3$

  (b).$2$

  (c).$1$

  (d).$6$
  Solution:
  $2(2k - 1) = k + (2k + 1) \implies 4k - 2 = 3k + 1 \implies k = 3$.
  Answer: (a) $3$
• 1 Mark Q17. The $4^{\text{th}}$ term from the end of the AP: $-11, -8, -5, \dots, 49$ is:
  (a).$40$

  (b).$37$

  (c).$43$

  (d).$34$
  Solution:
  $l = 49, d = 3$. $4^{\text{th}}$ term from end $= l - (4-1)d = 49 - 3(3) = 40$.
  Answer: (a) $40$
• 1 Mark Q18. The total number of terms in the AP $7, 13, 19, \dots, 205$ is:
  (a).$34$

  (b).$33$

  (c).$35$

  (d).$36$
  Solution:
  $205 = 7 + (n-1)6 \implies 198 = 6(n-1) \implies n-1 = 33 \implies n = 34$.
  Answer: (a) $34$
• 1 Mark Q19. If the sum of three numbers in AP is $27$ and their product is $504$, then the common difference $d$ can be:
  (a).$\pm 7$

  (b).$\pm 5$

  (c).$\pm 3$

  (d).$\pm 9$
  Solution:
  Let numbers be $a-d, a, a+d$. Sum $= 3a = 27 \implies a = 9$.
  Product $(9-d)(9)(9+d) = 504 \implies 81 - d^2 = 56 \implies d^2 = 25 \implies d = \pm 5$.
  Answer: (b) $\pm 5$ (Wait, option (a) was listed as $\pm 7$, let's fix the option display order or match correct answer: $\pm 5$ is option (b)).
  Answer: (b) $\pm 5$
• 1 Mark Q20. The common difference of the AP $\frac{1}{2b}, \frac{1 - 6b}{2b}, \frac{1 - 12b}{2b}, \dots$ is:
  (a).$-3$

  (b).$-3b$

  (c).$3$

  (d).$-6$
  Solution:
  $d = \frac{1 - 6b}{2b} - \frac{1}{2b} = \frac{-6b}{2b} = -3$.
  Answer: (a) $-3$
• 1 Mark Q21. The middle term of the AP $213, 205, 197, \dots, 37$ is:
  (a).$125$

  (b).$117$

  (c).$133$

  (d).$109$
  Solution:
  $37 = 213 + (n-1)(-8) \implies -176 = -8(n-1) \implies n = 23$.
  Middle term is $\frac{23+1}{2} = 12^{\text{th}}$ term: $a_{12} = 213 + 11(-8) = 125$.
  Answer: (a) $125$
• 1 Mark Q22. If the sum of the first $n$ terms of two APs are in the ratio $(3n + 8) : (7n + 15)$, then the ratio of their $12^{\text{th}}$ terms is:
  (a).$7 : 16$

  (b).$4 : 9$

  (c).$3 : 7$

  (d).$1 : 2$
  Solution:
  Replace $n$ with $2(12) - 1 = 23$: $\frac{3(23) + 8}{7(23) + 15} = \frac{77}{176} = \frac{7}{16}$.
  Answer: (a) $7 : 16$
• 1 Mark Q23. If $\log_2 2, \log_2 (2^x - 1),$ and $\log_2 (2^x + 3)$ are in AP, then the value of $x$ is:
  (a).$\log_2 5$

  (b).$1$

  (c).$\log_5 2$

  (d).$0$
  Solution:
  $2\log_2(2^x - 1) = \log_2 2 + \log_2(2^x + 3) \implies (2^x - 1)^2 = 2(2^x + 3)$.
  Let $2^x = y$: $(y-1)^2 = 2(y+3) \implies y^2 - 4y - 5 = 0 \implies y = 5$ (since $y > 1$).
  $2^x = 5 \implies x = \log_2 5$.
  Answer: (a) $\log_2 5$
• 1 Mark Q25. If $a_1, a_2, a_3, \dots, a_n$ are in AP with common difference $d$, then the sum of $\sin d [\sec a_1 \sec a_2 + \sec a_2 \sec a_3 + \dots + \sec a_{n-1} \sec a_n]$ is:
  (a).$\tan a_n - \tan a_1$

  (b).$\tan a_n + \tan a_1$

  (c).$\sin a_n - \sin a_1$

  (d).$\cos a_1 - \cos a_n$
  Solution:
  Each term can be rewritten using $\sin d = \sin(a_{k+1} - a_k)$:
  $\sin d \sec a_k \sec a_{k+1} = \frac{\sin(a_{k+1} - a_k)}{\cos a_k \cos a_{k+1}} = \frac{\sin a_{k+1}\cos a_k - \cos a_{k+1}\sin a_k}{\cos a_k \cos a_{k+1}} = \tan a_{k+1} - \tan a_k$.
  Summing from $k=1$ to $n-1$ yields telescoping sums: $(\tan a_2 - \tan a_1) + (\tan a_3 - \tan a_2) + \dots + (\tan a_n - \tan a_{n-1}) = \tan a_n - \tan a_1$.
  Answer: (a) $\tan a_n - \tan a_1$
• 1 Mark Q26. If $a_1, a_2, a_3, \dots$ is an AP such that $a_1 + a_5 + a_{10} + a_{15} + a_{20} + a_{24} = 225$, then $S_{24}$ is equal to:
  (a).$900$

  (b).$450$

  (c).$675$

  (d).$1800$
  Solution:
  In an AP, the sum of terms equidistant from the beginning and end is equal: $a_1 + a_{24} = a_5 + a_{20} = a_{10} + a_{15}$.
  Given sum $= 3(a_1 + a_{24}) = 225 \implies a_1 + a_{24} = 75$.
  $S_{24} = \frac{24}{2}(a_1 + a_{24}) = 12 \times 75 = 900$.
  Answer: (a) $900$
• 1 Mark Q27. The sum of the first $n$ terms of the series $1 + (1 + 2) + (1 + 2 + 3) + \dots$ is:
  (a).$\frac{n(n+1)(n+2)}{6}$

  (b).$\frac{n(n+1)(2n+1)}{6}$

  (c).$\frac{n^2(n+1)^2}{4}$

  (d).$\frac{n(n+1)}{2}$
  Solution:
  The $k^{\text{th}}$ term is $T_k = \frac{k(k+1)}{2} = \frac{1}{2}(k^2 + k)$.
  $S_n = \sum_{k=1}^n T_k = \frac{1}{2} \left[ \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right] = \frac{n(n+1)(n+2)}{6}$.
  Answer: (a) $\frac{n(n+1)(n+2)}{6}$
• 1 Mark Q28. If the $p^{\text{th}}, q^{\text{th}},$ and $r^{\text{th}}$ terms of an AP are $a, b, c$ respectively, then $a(q - r) + b(r - p) + c(p - q)$ is equal to:
  (a).$0$

  (b).$1$

  (c).$a + b + c$

  (d).$p + q + r$
  Solution:
  Let the first term be $A$ and common difference be $D$. Then $a = A + (p-1)D$, $b = A + (q-1)D$, $c = A + (r-1)D$.
  Substituting these into $a(q-r) + b(r-p) + c(p-q)$ yields $0$ due to cyclic cancellation of terms containing $A$ and $D$.
  Answer: (a) $0$
• 1 Mark Q29. How many three-digit numbers are divisible by $6$?
  (a).$150$

  (b).$149$

  (c).$151$

  (d).$166$
  Solution:
  First 3-digit multiple of $6$ is $102$, and the last is $996$.
  $996 = 102 + (n-1)6 \implies 894 = 6(n-1) \implies n-1 = 149 \implies n = 150$.
  Answer: (a) $150$
• 1 Mark Q30. If $S_1, S_2, S_3$ are the sums of $n$ terms of three APs whose first terms are $1, 2, 3$ and common differences are $1, 3, 5$ respectively, then $S_1 + S_3$ equals:
  (a).$2 S_2$

  (b).$S_2$

  (c).$3 S_2$

  (d).$4 S_2$
  Solution:
  $S_1 = \frac{n}{2}[2(1) + (n-1)1] = \frac{n}{2}(n+1)$, $S_2 = \frac{n}{2}[2(2) + (n-1)3] = \frac{n}{2}(3n+1)$, $S_3 = \frac{n}{2}[2(3) + (n-1)5] = \frac{n}{2}(5n+1)$.
  $S_1 + S_3 = \frac{n}{2}(6n+2) = 2 \cdot \frac{n}{2}(3n+1) = 2S_2$.
  Answer: (a) $2 S_2$
• 1 Mark Q31. If $a, b, c$ are in AP, then the value of $(a + 2b - c)(2b + c - a)(c + a - b)$ is:
  (a).$4abc$

  (b).$abc$

  (c).$8abc$

  (d).$2abc$
  Solution:
  Since $a, b, c$ are in AP, $2b = a + c$.
  Substitute $2b = a+c$: $(a + a + c - c)(a + c + c - a)(c + a - \frac{a+c}{2}) = (2a)(2c)(b) = 4abc$.
  Answer: (a) $4abc$
• 1 Mark Q32. The interior angles of a convex polygon are in AP. The smallest angle is $120^\circ$ and the common difference is $5^\circ$. The number of sides of the polygon is:
  (a).$9$

  (b).$16$

  (c).$9$ or $16$

  (d).$12$
  Solution:
  $\frac{n}{2}[2(120) + (n-1)5] = (n-2)180 \implies 5n^2 - 125n + 720 = 0 \implies (n-9)(n-16) = 0$.
  $n=16$ yields an angle of $195^\circ$, which is invalid for a convex polygon. Thus, $n=9$.
  Answer: (a) $9$
• 1 Mark Q33. If the ratio of the $m^{\text{th}}$ and $n^{\text{th}}$ terms of an AP is $(2m - 1) : (2n - 1)$, then the ratio of the sum of its first $m$ and $n$ terms is:
  (a).$m^2 : n^2$

  (b).$m : n$

  (c).$(2m + 1) : (2n + 1)$

  (d).$m^3 : n^3$
  Solution:
  Using standard AP relations, if the ratio of $T_m / T_n = (2m-1)/(2n-1)$, the ratio of sum of $m$ and $n$ terms is $m^2 : n^2$.
  Answer: (a) $m^2 : n^2$
• 1 Mark Q34. The sum of all $2$-digit numbers which are NOT divisible by $3$ is:
  (a).$3240$

  (b).$4905$

  (c).$1665$

  (d).$3300$
  Solution:
  Sum of all 2-digit numbers $= 4905$. Sum of 2-digit numbers divisible by $3$ ($12$ to $99$, $n=30$) $= 1665$.
  Required sum $= 4905 - 1665 = 3240$.
  Answer: (a) $3240$
• 1 Mark Q35. If $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in AP, then $\frac{b+a}{b-a} + \frac{b+c}{b-c}$ is equal to:
  (a).$2$

  (b).$1$

  (c).$0$

  (d).$-1$
  Solution:
  $\frac{2}{b} = \frac{1}{a} + \frac{1}{c} \implies b = \frac{2ac}{a+c}$. Substituting this into the expression yields $2$.
  Answer: (a) $2$
• 1 Mark Q36. A ladder has rungs $25\text{ cm}$ apart. The rungs decrease uniformly in length from $45\text{ cm}$ at the bottom to $25\text{ cm}$ at the top. If the top and bottom rungs are $2.5\text{ m}$ apart, what length of wood is required for the rungs?
  (a).$385\text{ cm}$

  (b).$350\text{ cm}$

  (c).$410\text{ cm}$

  (d).$360\text{ cm}$
  Solution:
  Number of spaces $= \frac{250}{25} = 10 \implies n = 11$ rungs.
  Total length $= \frac{11}{2}(45 + 25) = 385\text{ cm}$.
  Answer: (a) $385\text{ cm}$
• 1 Mark Q37. If $a, b, c$ are in AP, then $a^2(b + c), b^2(c + a), c^2(a + b)$ are:
  (a).in AP

  (b).not in AP

  (c).in GP

  (d).all equal to $0$
  Solution:
  Testing with values $a=1, b=2, c=3$, the terms are $5, 16, 27$, which form an AP with a common difference of $11$.
  Answer: (a) in AP
• 1 Mark Q38. In an AP, if $S_n = n(4n + 1)$, then the $15^{\text{th}}$ term is:
  (a).$117$

  (b).$115$

  (c).$121$

  (d).$109$
  Solution:
  $a_{15} = S_{15} - S_{14} = 15(61) - 14(57) = 915 - 798 = 117$.
  Answer: (a) $117$
• 1 Mark Q39. If $1 + 6 + 11 + 16 + \dots + x = 148$, then the value of $x$ is:
  (a).$36$

  (b).$41$

  (c).$31$

  (d).$46$
  Solution:
  $S_n = \frac{n}{2}(5n - 3) = 148 \implies 5n^2 - 3n - 296 = 0 \implies n = 8$.
  $x = a_8 = 1 + 7(5) = 36$.
  Answer: (a) $36$
• 1 Mark Q40. If the sum of first $n$ even natural numbers is equal to $k$ times the sum of first $n$ odd natural numbers, then $k$ equals:
  (a).$\frac{n + 1}{n}$

  (b).$\frac{n}{n + 1}$

  (c).$\frac{n + 1}{2n}$

  (d).$\frac{n - 1}{n}$
  Solution:
  Sum of even natural numbers $= n(n+1)$, sum of odd natural numbers $= n^2$.
  $n(n+1) = k \cdot n^2 \implies k = \frac{n+1}{n}$.
  Answer: (a) $\frac{n + 1}{n}$
• 1 Mark Q41. If the terms $a, b, c, d$ form an AP, then $a - 4b + 6c - 4d + e$ equals zero when $e$ is:
  (a).the $5^{\text{th}}$ term of the AP

  (b).any arbitrary constant

  (c).$0$

  (d).the sum of $a$ and $d$
  Solution:
  Using binomial coefficients $1, -4, 6, -4, 1$ for an AP, the expression vanishes when $e$ is the $5^{\text{th}}$ term of the AP.
  Answer: (a) the $5^{\text{th}}$ term of the AP
• 1 Mark Q42. The $n^{\text{th}}$ term of an AP is given by $a_n = 3 + 4n$. The sum of the first $15$ terms is:
  (a).$525$

  (b).$465$

  (c).$555$

  (d).$495$
  Solution:
  $a_1 = 7$, $a_{15} = 63$. $S_{15} = \frac{15}{2}(7 + 63) = 525$.
  Answer: (a) $525$
• 1 Mark Q43. If $18, a, b, -3$ are in AP, then $a + b$ is equal to:
  (a).$15$

  (b).$19$

  (c).$21$

  (d).$11$
  Solution:
  Sum of extremes equals the sum of interior terms in a 4-term AP: $18 + (-3) = a + b \implies a + b = 15$.
  Answer: (a) $15$
• 1 Mark Q44. The sum of all natural numbers between $100$ and $1000$ which are multiples of $5$ is:
  (a).$98450$

  (b).$98550$

  (c).$99450$

  (d).$97450$
  Solution:
  Multiples from $105$ to $995$: $n = 179$. Sum $= \frac{179}{2}(105 + 995) = 98450$.
  Answer: (a) $98450$
• 1 Mark Q45. If the $9^{\text{th}}$ term of an AP is zero, then the ratio of its $29^{\text{th}}$ term to its $19^{\text{th}}$ term is:
  (a).$2 : 1$

  (b).$3 : 1$

  (c).$1 : 2$

  (d).$3 : 2$
  Solution:
  $a + 8d = 0 \implies a = -8d$. Ratio $= \frac{a+28d}{a+18d} = \frac{20d}{10d} = 2:1$.
  Answer: (a) $2 : 1$
• 1 Mark Q46. If $b + c - a, c + a - b, a + b - c$ are in AP, then $a, b, c$ are in:
  (a).AP

  (b).GP

  (c).HP

  (d).None of these
  Solution:
  $2(c + a - b) = (b + c - a) + (a + b - c) \implies 2c + 2a - 2b = 2b \implies a + c = 2b$.
  Answer: (a) AP
• 1 Mark Q47. The value of $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{n(n+1)}$ is:
  (a).$\frac{n}{n+1}$

  (b).$\frac{1}{n+1}$

  (c).$\frac{n+1}{n}$

  (d).$\frac{2n}{n+1}$
  Solution:
  Using telescoping sums: $\sum_{k=1}^n \left(\frac{1}{k} - \frac{1}{k+1}\right) = 1 - \frac{1}{n+1} = \frac{n}{n+1}$.
  Answer: (a) $\frac{n}{n+1}$
• 1 Mark Q48. The common difference of an AP in which $a_{21} - a_7 = 84$ is:
  (a).$6$

  (b).$8$

  (c).$4$

  (d).$12$
  Solution:
  $(a + 20d) - (a + 6d) = 84 \implies 14d = 84 \implies d = 6$.
  Answer: (a) $6$
• 1 Mark Q49. In an AP, if $a = 1$, $a_n = 20$, and $S_n = 399$, then $n$ is equal to:
  (a).$38$

  (b).$42$

  (c).$36$

  (d).$40$
  Solution:
  $S_n = \frac{n}{2}(a + a_n) \implies 399 = \frac{n}{2}(1 + 20) \implies 21n = 798 \implies n = 38$.
  Answer: (a) $38$
• 1 Mark Q50. If $15^{\text{th}}$ term of an AP is $0$, then its $21^{\text{st}}$ term and $11^{\text{th}}$ term are in the ratio:
  (a).$3 : -2$

  (b).$2 : -3$

  (c).$3 : 2$

  (d).$-3 : 2$
  Solution:
  $a + 14d = 0 \implies a = -14d$. Ratio $= \frac{a+20d}{a+10d} = \frac{6d}{-4d} = 3:-2$.
  Answer: (a) $3 : -2$

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