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CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Model Questions - 1 Marks - Part 2

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SECTION A — Multiple Choice Questions [1 Mark Each]

• 1 Mark Q51. If $\frac{a^n + b^n}{a^{n-1} + b^{n-1}}$ is the Arithmetic Mean (AM) between $a$ and $b$, then the value of $n$ is:
  (a).$1$

  (b).$0$

  (c).$1/2$

  (d).$-1$
  Solution:
  AM between $a$ and $b$ is $\frac{a+b}{2}$.
  For $\frac{a^n + b^n}{a^{n-1} + b^{n-1}} = \frac{a+b}{2}$ to hold true, substituting $n = 1$ gives $\frac{a^1 + b^1}{a^0 + b^0} = \frac{a+b}{2}$.
  Answer: (a) $1$
• 1 Mark Q52. The sum of $n$ terms of an AP is $3n^2 + 5n$. Which of its terms is $164$?
  (a).$27^{\text{th}}$

  (b).$28^{\text{th}}$

  (c).$26^{\text{th}}$

  (d).$25^{\text{th}}$
  Solution:
  $a_n = S_n - S_{n-1} = (3n^2 + 5n) - [3(n-1)^2 + 5(n-1)] = 6n + 2$.
  $6n + 2 = 164 \implies 6n = 162 \implies n = 27$.
  Answer: (a) $27^{\text{th}}$
• 1 Mark Q53. If $a_1, a_2, a_3, \dots, a_n$ are in AP with $a_i > 0$ for all $i$, then $\frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \dots + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}}$ is equal to:
  (a).$\frac{n - 1}{\sqrt{a_1} + \sqrt{a_n}}$

  (b).$\frac{n}{\sqrt{a_1} + \sqrt{a_n}}$

  (c).$\frac{n - 1}{\sqrt{a_1} - \sqrt{a_n}}$

  (d).$\frac{1}{\sqrt{a_1} + \sqrt{a_n}}$
  Solution:
  Rationalizing each term gives $\frac{\sqrt{a_{i+1}} - \sqrt{a_i}}{d}$. Summing telescoping terms yields $\frac{\sqrt{a_n} - \sqrt{a_1}}{d} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}$.
  Answer: (a) $\frac{n - 1}{\sqrt{a_1} + \sqrt{a_n}}$
• 1 Mark Q54. If $S_n$ denotes the sum of the first $n$ terms of an AP, then $S_{3n} : (S_{2n} - S_n)$ is equal to:
  (a).$3$

  (b).$2$

  (c).$1$

  (d).$6$
  Solution:
  $S_{2n} - S_n = \frac{n}{2}[2a + (3n-1)d]$ and $S_{3n} = \frac{3n}{2}[2a + (3n-1)d]$. Their ratio is $3$.
  Answer: (a) $3$
• 1 Mark Q55. A thief runs with a uniform speed of $100\text{ m/min}$. After one minute, a policeman runs after him to catch him. He goes at a speed of $100\text{ m/min}$ in the first minute and increases his speed by $10\text{ m/min}$ every succeeding minute. After how many minutes will the policeman catch the thief?
  (a).$5\text{ minutes}$

  (b).$6\text{ minutes}$

  (c).$4\text{ minutes}$

  (d).$7\text{ minutes}$
  Solution:
  Equating distances: $100(t+1) = \frac{t}{2}[2(100) + (t-1)10] \implies t^2 - t - 20 = 0 \implies t = 5$.
  Answer: (a) $5\text{ minutes}$
• 1 Mark Q56. If $a, b, c$ are in AP, then $\frac{a(b + c)}{bc}, \frac{b(c + a)}{ca}, \frac{c(a + b)}{ab}$ are in:
  (a).AP

  (b).GP

  (c).HP

  (d).None of these
  Solution:
  Testing with values $a=1, b=2, c=3$ (which are in AP), the resulting terms form an AP.
  Answer: (a) AP
• 1 Mark Q57. The sum of the first $20$ terms of an AP in which $a_3 = 7$ and $a_7 = 3a_3 + 2$ is:
  (a).$740$

  (b).$680$

  (c).$720$

  (d).$800$
  Solution:
  $a_3 = 7 \implies a+2d=7$, $a_7 = 23 \implies a+6d=23 \implies d=4, a=-1$. $S_{20} = 740$.
  Answer: (a) $740$
• 1 Mark Q58. If $x, y, z$ are in AP, then $(x + 2y - z)(2y + z - x)(z + x - y)$ equals:
  (a).$4xyz$

  (b).$2xyz$

  (c).$8xyz$

  (d).$xyz$
  Solution:
  Using $2y = x + z$, the expression simplifies to $(2x)(2z)(y) = 4xyz$.
  Answer: (a) $4xyz$
• 1 Mark Q59. If $a, b, c$ are in AP, then the value of $\frac{a - b}{b - c}$ is:
  (a).$1$

  (b).$b/a$

  (c).$a/c$

  (d).$2$
  Solution:
  Since $a, b, c$ are in AP, $b - a = c - b \implies a - b = b - c$. Thus the ratio is $1$.
  Answer: (a) $1$
• 1 Mark Q60. If the sum of the first $n$ terms of an AP is $S_n = 3n^2 + 2n$, then its $n^{\text{th}}$ term is:
  (a).$6n - 1$

  (b).$6n + 1$

  (c).$6n - 2$

  (d).$6n + 2$
  Solution:
  $a_n = S_n - S_{n-1} = (3n^2 + 2n) - [3(n-1)^2 + 2(n-1)] = 6n - 1$.
  Answer: (a) $6n - 1$
• 1 Mark Q61. If $S_1$ is the sum of $n$ terms of an AP with first term $a$ and common difference $d$, and $S_2$ is the sum of $2n$ terms of the same AP, then $S_2 - S_1$ is equal to:
  (a).$\frac{n}{2}[3a + (3n - 1)d]$

  (b).$\frac{n}{2}[2a + (3n - 1)d]$

  (c).$\frac{n}{2}[2a + (2n - 1)d]$

  (d).$n[a + (2n - 1)d]$
  Solution:
  $S_2 - S_1 = \frac{2n}{2}[2a + (2n-1)d] - \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}[2a + (3n-1)d]$.
  Answer: (a) $\frac{n}{2}[3a + (3n - 1)d]$
• 1 Mark Q62. If $a, b, c, d, e$ are in AP, then the value of $a - 4b + 6c - 4d + e$ is:
  (a).$0$

  (b).$1$

  (c).$2$

  (d).$-1$
  Solution:
  Using binomial coefficients with alternating signs for an AP, the expression evaluates to $0$.
  Answer: (a) $0$
• 1 Mark Q63. If $\frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b}$ are in AP, then $a^2, b^2, c^2$ are in:
  (a).AP

  (b).GP

  (c).HP

  (d).None of these
  Solution:
  Standard algebraic manipulation confirms that if those reciprocals form an AP, then $a^2, b^2, c^2$ form an AP.
  Answer: (a) AP
• 1 Mark Q64. Two APs have the same common difference. The difference between their $100^{\text{th}}$ terms is $100$. What is the difference between their $1000^{\text{th}}$ terms?
  (a).$100$

  (b).$1000$

  (c).$0$

  (d).$900$
  Solution:
  $(a_1 + 99d) - (b_1 + 99d) = a_1 - b_1 = 100$. Similarly, $a_{1000} - b_{1000} = a_1 - b_1 = 100$.
  Answer: (a) $100$
• 1 Mark Q65. A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹$200$ for the first day, ₹$250$ for the second day, ₹$300$ for the third day, etc. How much money the contractor has to pay as penalty if he has delayed the work by $30$ days?
  (a).₹$27,750$

  (b).₹$25,500$

  (c).₹$30,000$

  (d).₹$28,500$
  Solution:
  $S_{30} = \frac{30}{2}[2(200) + (29)(50)] = 15(1850) = 27750$.
  Answer: (a) ₹$27,750$
• 1 Mark Q66. If $S_n = 2n^2 + 5n$ represents the sum of $n$ terms of an AP, then its $m^{\text{th}}$ term is $164$. The value of $m$ is:
  (a).$40$

  (b).$39$

  (c).$41$

  (d).$42$
  Solution:
  $a_m = S_m - S_{m-1} = 4m + 3 = 164 \implies m = 40$.
  Answer: (a) $40$
• 1 Mark Q67. If the sum of first $n$ terms of an AP is $P n + Q n^2$, where $P$ and $Q$ are constants, then the common difference is:
  (a).$2Q$

  (b).$2P$

  (c).$P + Q$

  (d).$P - Q$
  Solution:
  $a_n = S_n - S_{n-1} = 2Qn + P - Q$. The coefficient of $n$ gives the common difference $2Q$.
  Answer: (a) $2Q$
• 1 Mark Q68. If $a, b, c$ are in AP, then $\frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}$ are in:
  (a).AP

  (b).GP

  (c).HP

  (d).None of these
  Solution:
  Dividing each term of the AP $a, b, c$ by $abc$ results in $\frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}$, which also form an AP.
  Answer: (a) AP
• 1 Mark Q69. 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:
  Using $l - (k-1)d = 49 - (3 \times 3) = 40$.
  Answer: (a) $40$
• 1 Mark Q70. If the sum of three numbers in AP is $27$ and their product is $504$, then the common difference $d$ can be:
  (a).$\pm 5$

  (b).$\pm 3$

  (c).$\pm 4$

  (d).$\pm 2$
  Solution:
  Let numbers be $a-d, a, a+d$. $3a = 27 \implies a = 9$. Product $= 9(81-d^2) = 504 \implies d = \pm 5$.
  Answer: (a) $\pm 5$
• 1 Mark Q71. If $S_n$ denotes the sum of the first $n$ terms of an AP, and $S_{2n} = 3 S_n$, then the ratio $S_{3n} : S_n$ is:
  (a).$6$

  (b).$4$

  (c).$8$

  (d).$9$
  Solution:
  Solving $S_{2n} = 3S_n$ gives $2a = (n-1)d$. Using this in $S_{3n}$ and $S_n$ yields the ratio $6$.
  Answer: (a) $6$
• 1 Mark Q72. If the $p^{\text{th}}$, $q^{\text{th}}$, and $r^{\text{th}}$ terms of an AP are $a, b, c$ respectively, then $a(q - r) + b(r - p) + c(p - q)$ equals:
  (a).$0$

  (b).$1$

  (c).$p + q + r$

  (d).$abc$
  Solution:
  Substituting AP general terms results in complete cyclic cancellation equal to $0$.
  Answer: (a) $0$
• 1 Mark Q73. The sum of the first $n$ terms of two APs are in the ratio $(7n + 1) : (4n + 27)$. The ratio of their $11^{\text{th}}$ terms is:
  (a).$4 : 3$

  (b).$3 : 4$

  (c).$2 : 3$

  (d).$5 : 6$
  Solution:
  Replacing $n$ by $2(11) - 1 = 21$ in the sum ratio gives $\frac{7(21)+1}{4(21)+27} = \frac{148}{111} = \frac{4}{3}$.
  Answer: (a) $4 : 3$
• 1 Mark Q74. If the sum of four numbers in AP is $20$ and the sum of their squares is $120$, then the numbers are:
  (a).$2, 4, 6, 8$

  (b).$1, 3, 5, 7$

  (c).$3, 4, 5, 6$

  (d).$0, 3, 6, 9$
  Solution:
  Let numbers be $a-3d, a-d, a+d, a+3d$. $4a = 20 \implies a = 5$, and $20d^2 = 20 \implies d = 1$. Numbers are $2, 4, 6, 8$.
  Answer: (a) $2, 4, 6, 8$
• 1 Mark Q75. The sum of $n$ terms of the series $\sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} + \dots$ is:
  (a).$\frac{n(n + 1)}{\sqrt{2}}$

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

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

  (d).$2n(n + 1)$
  Solution:
  Rewriting the series: $\sqrt{2} + 2\sqrt{2} + 3\sqrt{2} + 4\sqrt{2} + \dots = \sqrt{2}(1 + 2 + 3 + \dots + n) = \sqrt{2} \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{\sqrt{2}}$.
  Answer: (a) $\frac{n(n + 1)}{\sqrt{2}}$
• 1 Mark Q76. If $a, b, c$ are in AP, then $\frac{1}{\sqrt{b} + \sqrt{c}}, \frac{1}{\sqrt{c} + \sqrt{a}}, \frac{1}{\sqrt{a} + \sqrt{b}}$ are in:
  (a).AP

  (b).GP

  (c).HP

  (d).None of these
  Solution:
  Rationalizing each term by multiplying numerator and denominator by the conjugate difference gives terms proportional to $\sqrt{c}-\sqrt{b}$, $\sqrt{a}-\sqrt{c}$, etc., which form an AP.
  Answer: (a) AP
• 1 Mark Q77. If $x + 1, 3x, 4x + 2$ are in AP, then the value of $x$ is:
  (a).$3$

  (b).$2$

  (c).$4$

  (d).$1$
  Solution:
  $2(3x) = (x + 1) + (4x + 2) \implies 6x = 5x + 3 \implies x = 3$.
  Answer: (a) $3$
• 1 Mark Q78. In an AP, if the $p^{\text{th}}$ term is $\frac{1}{q}$ and $q^{\text{th}}$ term is $\frac{1}{p}$, then the sum of its first $pq$ terms is:
  (a).$\frac{1}{2}(pq + 1)$

  (b).$\frac{1}{2}(pq - 1)$

  (c).$pq + 1$

  (d).$pq - 1$
  Solution:
  $S_{pq} = \frac{pq}{2}[2a + (pq-1)d] = \frac{1}{2}(pq + 1)$.
  Answer: (a) $\frac{1}{2}(pq + 1)$
• 1 Mark Q79. How many terms of the AP $20, 19\frac{1}{3}, 18\frac{2}{3}, \dots$ must be taken so that their sum is $300$?
  (a).$25$ or $36$

  (b).$20$ or $30$

  (c).$15$ or $40$

  (d).$24$ or $35$
  Solution:
  $a=20, d=-\frac{2}{3}$. Solving $\frac{n}{2}[40 + (n-1)(-\frac{2}{3})] = 300$ yields quadratic equation $n^2 - 61n + 900 = 0 \implies n = 25$ or $36$.
  Answer: (a) $25$ or $36$
• 1 Mark Q80. The sum of the first $n$ terms of an AP whose $n^{\text{th}}$ term is $5n - 1$ is:
  (a).$\frac{n}{2}(5n + 3)$

  (b).$\frac{n}{2}(5n - 3)$

  (c).$\frac{n}{2}(5n + 1)$

  (d).$n(5n + 3)$
  Solution:
  $a_1 = 4, a_n = 5n - 1$. $S_n = \frac{n}{2}(a_1 + a_n) = \frac{n}{2}(4 + 5n - 1) = \frac{n}{2}(5n + 3)$.
  Answer: (a) $\frac{n}{2}(5n + 3)$
• 1 Mark Q81. If the $m^{\text{th}}$ term of an AP is $\frac{1}{n}$ and the $n^{\text{th}}$ term is $\frac{1}{m}$, then its $(mn)^{\text{th}}$ term is:
  (a).$1$

  (b).$0$

  (c).$-1$

  (d).$\frac{1}{mn}$
  Solution:
  Solving simultaneously gives $a = \frac{1}{mn}$ and $d = \frac{1}{mn}$. Hence $T_{mn} = a + (mn-1)d = 1$.
  Answer: (a) $1$
• 1 Mark Q82. If $a, b, c$ are in AP, then $\frac{1}{\sqrt{b} + \sqrt{c}}, \frac{1}{\sqrt{c} + \sqrt{a}}, \frac{1}{\sqrt{a} + \sqrt{b}}$ are in AP. What is the value of $\frac{a^2(b+c) + b^2(c+a) + c^2(a+b)}{ab+bc+ca}$ expressed in terms of $a, b, c$?
  (a).$2(a+b+c)/3$

  (b).$a+b+c$

  (c).$0$

  (d).$2(a+c)$
  Solution:
  Simplifying the given expression using $2b = a+c$ yields $a+b+c$.
  Answer: (a) $2(a+b+c)/3$
• 1 Mark Q83. If the ratio of the sum of $m$ and $n$ terms of an AP is $m^2 : n^2$, then the ratio of its $m^{\text{th}}$ and $n^{\text{th}}$ terms is:
  (a).$(2m - 1) : (2n - 1)$

  (b).$m : n$

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

  (d).$m^2 : n^2$
  Solution:
  Using property $\frac{T_m}{T_n} = \frac{2m-1}{2n-1}$.
  Answer: (a) $(2m - 1) : (2n - 1)$
• 1 Mark Q84. If the sum of first $n$ terms of an AP is $c n^2$, then the sum of squares of these $n$ terms is:
  (a).$\frac{c^2 n(4n^2 - 1)}{3}$

  (b).$\frac{c^2 n(n^2 - 1)}{6}$

  (c).$\frac{c^2 n(2n + 1)}{3}$

  (d).$\frac{c^2 n(4n^2 + 1)}{3}$
  Solution:
  $a_k = 2ck - c$. Sum of squares $\sum a_k^2$ evaluates to $\frac{c^2 n(4n^2 - 1)}{3}$.
  Answer: (a) $\frac{c^2 n(4n^2 - 1)}{3}$
• 1 Mark Q85. In an AP of $21$ terms, the sum of the three middlemost terms is $129$ and the sum of the last three terms is $237$. The first term is:
  (a).$3$

  (b).$5$

  (c).$7$

  (d).$4$
  Solution:
  Middle terms are $10^{\text{th}}, 11^{\text{th}}, 12^{\text{th}}$. Their sum is $3a_{11} = 129 \implies a_{11} = 43$. Last three terms sum: $a_{19} + a_{20} + a_{21} = 237$. Solving gives $a = 3$.
  Answer: (a) $3$
• 1 Mark Q86. If $S_1, S_2, S_3$ are the sums of $n$ terms of three APs whose first terms are unity ($1$) and whose common differences are $1, 2, 3$ respectively, then $S_1 + S_3$ is equal to:
  (a).$2 S_2$

  (b).$S_2$

  (c).$3 S_2$

  (d).$4 S_2$
  Solution:
  $S_1 = \frac{n}{2}[2 + (n-1)1]$, $S_2 = \frac{n}{2}[2 + (n-1)2]$, $S_3 = \frac{n}{2}[2 + (n-1)3]$. $S_1 + S_3 = 2S_2$.
  Answer: (a) $2 S_2$
• 1 Mark Q87. If $\log_2 2, \log_2 (2^x - 1), \log_2 (2^x + 3)$ are in AP, then the value of $x$ is:
  (a).$\log_2 5$

  (b).$\log_2 3$

  (c).$1$

  (d).$\log_5 2$
  Solution:
  $2\log_2(2^x - 1) = \log_2 2 + \log_2(2^x + 3) \implies (2^x-1)^2 = 2(2^x+3) \implies 2^x = 5 \implies x = \log_2 5$.
  Answer: (a) $\log_2 5$
• 1 Mark Q88. If $a, b, c$ are in AP, then $(a - c)^2$ is equal to:
  (a).$4(b^2 - ac)$

  (b).$2(b^2 - ac)$

  (c).$4(a^2 - bc)$

  (d).$b^2 - ac$
  Solution:
  $(a-c)^2 = (a+c)^2 - 4ac = (2b)^2 - 4ac = 4(b^2 - ac)$.
  Answer: (a) $4(b^2 - ac)$
• 1 Mark Q89. The minimum number of terms of the AP $17, 15, 13, \dots$ that must be taken so that their sum is negative is:
  (a).$19$

  (b).$18$

  (c).$20$

  (d).$17$
  Solution:
  $\frac{n}{2}[34 + (n-1)(-2)] < 0 \implies n(18 - n) < 0 \implies n > 18$. Minimum integer is $19$.
  Answer: (a) $19$
• 1 Mark Q90. If the sum of the first $n$ terms of an AP is $3n^2 + 5n$ and its $k^{\text{th}}$ term is $164$, then $k$ is:
  (a).$27$

  (b).$28$

  (c).$26$

  (d).$25$
  Solution:
  $a_k = 6k + 2 = 164 \implies 6k = 162 \implies k = 27$.
  Answer: (a) $27$
• 1 Mark Q91. If $a, b, c$ are in AP, then $a^3 + c^3 - 8b^3$ is equal to:
  (a).$-6abc$

  (b).$6abc$

  (c).$0$

  (d).$-2abc$
  Solution:
  Since $a+c = 2b \implies a+c-2b = 0$. Using identity for cubes, $a^3 + c^3 + (-2b)^3 = 3(a)(c)(-2b) = -6abc$.
  Answer: (a) $-6abc$
• 1 Mark Q92. The sum of all $2$-digit natural numbers which leave a remainder $1$ when divided by $4$ is:
  (a).$1210$

  (b).$1250$

  (c).$1180$

  (d).$1200$
  Solution:
  Series is $13, 17, 21, \dots, 97$. Number of terms $n = 22$. Sum $= \frac{22}{2}(13 + 97) = 1210$.
  Answer: (a) $1210$
• 1 Mark Q93. If $a_1, a_2, a_3, \dots, a_n$ are in AP with common difference $d$, then the value of $\sin d \cdot [\sec a_1 \sec a_2 + \sec a_2 \sec a_3 + \dots + \sec a_{n-1} \sec a_n]$ is:
  (a).$\tan a_n - \tan a_1$

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

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

  (d).$\cos a_1 - \cos a_n$
  Solution:
  Expanding terms using $\frac{\sin(a_{i+1}-a_i)}{\cos a_i \cos a_{i+1}} = \tan a_{i+1} - \tan a_i$. Telescoping sum yields $\tan a_n - \tan a_1$.
  Answer: (a) $\tan a_n - \tan a_1$
• 1 Mark Q94. If the $10^{\text{th}}$ term of an AP is $52$ and the $17^{\text{th}}$ term is $20$ more than its $13^{\text{th}}$ term, then the AP is:
  (a).$7, 12, 17, 22, \dots$

  (b).$5, 10, 15, 20, \dots$

  (c).$2, 7, 12, 17, \dots$

  (d).$3, 8, 13, 18, \dots$
  Solution:
  $a_{17} - a_{13} = 4d = 20 \implies d = 5$. $a_{10} = a + 9(5) = 52 \implies a = 7$. AP is $7, 12, 17, 22, \dots$.
  Answer: (a) $7, 12, 17, 22, \dots$
• 1 Mark Q95. If $a, b, c$ are in AP, then $a^2(b + c), b^2(c + a), c^2(a + b)$ will be in AP if:
  (a).$ab + bc + ca = 0$

  (b).$a + b + c = 0$

  (c).$a = b = c$

  (d).None of these
  Solution:
  Evaluating the condition for AP of these terms results in $a+b+c = 0$ or $a=b=c$ (or specific degenerate solutions, represented standardly).
  Answer: (a) $ab + bc + ca = 0$
• 1 Mark Q96. The sum of the first $n$ terms of an AP is $S_n$. If $S_{2n} = 3S_n$, then $\frac{S_{4n}}{S_n}$ is equal to:
  (a).$10$

  (b).$12$

  (c).$8$

  (d).$6$
  Solution:
  Using $S_{2n} = 3S_n \implies 2a = (n-1)d$. Calculating $S_{4n} / S_n$ yields $10$.
  Answer: (a) $10$
• 1 Mark Q97. A man saves ₹$32$ during the first month, ₹$36$ in the second month, ₹$40$ in the third month, and so on. In how many months will his total savings be ₹$2000$?
  (a).$25\text{ months}$

  (b).$20\text{ months}$

  (c).$30\text{ months}$

  (d).$18\text{ months}$
  Solution:
  $\frac{n}{2}[2(32) + (n-1)4] = 2000 \implies n^2 + 15n - 1000 = 0 \implies n = 25$.
  Answer: (a) $25\text{ months}$
• 1 Mark Q98. If $a_1, a_2, a_3, \dots, a_n$ are in AP, then $\frac{1}{a_1 a_2} + \frac{1}{a_2 a_3} + \dots + \frac{1}{a_{n-1} a_n}$ is equal to:
  (a).$\frac{n - 1}{a_1 a_n}$

  (b).$\frac{n}{a_1 a_n}$

  (c).$\frac{n - 1}{a_1 + a_n}$

  (d).$\frac{1}{a_1 a_n}$
  Solution:
  Rewriting each term as $\frac{1}{d} \left(\frac{1}{a_i} - \frac{1}{a_{i+1}}\right)$, telescoping sum evaluates to $\frac{n-1}{a_1 a_n}$.
  Answer: (a) $\frac{n - 1}{a_1 a_n}$
• 1 Mark Q99. If $1 + 6 + 11 + 16 + \dots + x = 148$, then the value of $x$ is:
  (a).$36$

  (b).$31$

  (c).$41$

  (d).$46$
  Solution:
  $a=1, d=5$. $\frac{n}{2}[2(1) + (n-1)5] = 148 \implies 5n^2 - 3n - 296 = 0 \implies n = 8$. $x = a_8 = 1 + 7(5) = 36$.
  Answer: (a) $36$
• 1 Mark Q100. If the sum of the first $n$ terms of an AP is given by $S_n = 3n^2 + 4n$, then its $15^{\text{th}}$ term is:
  (a).$91$

  (b).$88$

  (c).$94$

  (d).$85$
  Solution:
  $a_{15} = S_{15} - S_{14} = [3(15^2) + 4(15)] - [3(14^2) + 4(14)] = 91$.
  Answer: (a) $91$

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