Practice Questions: Arithmetic Progression (AP)
Question 1: Find the $10^{\text{th}}$ term of the AP: $2, 4, 6, 8, \dots$
Solution: Here, the first term $a = 2$ and the common difference $d = 4 - 2 = 2$. The $n^{\text{th}}$ term is given by $a_n = a + (n-1)d$. For $n=10$, $a_{10} = 2 + (10-1)2 = 2 + 18 = 20$.
Question 2: What is the $15^{\text{th}}$ term of the AP: $1, 4, 7, 10, \dots$?
Solution: $a = 1$, $d = 3$. $a_{15} = 1 + (15-1)3 = 1 + 14(3) = 1 + 42 = 43$.
Question 3: Find the $20^{\text{th}}$ term of the sequence: $10, 8, 6, 4, \dots$
Solution: $a = 10$, $d = 8 - 10 = -2$. $a_{20} = 10 + 19(-2) = 10 - 38 = -28$.
Question 4: In an AP, the first term is $5$ and the fifth term is $17$. Find the common difference.
Solution: $a = 5$, $a_5 = 17$. Since $a_5 = a + 4d$, we have $5 + 4d = 17 \implies 4d = 12 \implies d = 3$.
Question 5: If the $3^{\text{rd}}$ term of an AP is $10$ and the $6^{\text{th}}$ term is $19$, find the first term $a$.
Solution: $a + 2d = 10$ and $a + 5d = 19$. Subtracting the first from the second gives $3d = 9 \implies d = 3$. Substituting $d$ back, $a + 6 = 10 \implies a = 4$.
Question 6: How many terms are there in the AP: $3, 6, 9, \dots, 30$?
Solution: $a = 3$, $d = 3$, $a_n = 30$. Using $a_n = a + (n-1)d \implies 30 = 3 + (n-1)3 \implies 27 = 3(n-1) \implies n-1 = 9 \implies n = 10$.
Question 7: Find the number of terms in the AP: $5, 10, 15, \dots, 100$.
Solution: $a = 5$, $d = 5$, $a_n = 100$. $100 = 5 + (n-1)5 \implies 95 = 5(n-1) \implies 19 = n-1 \implies n = 20$.
Question 8: How many terms are in the sequence $100, 95, 90, \dots, 5$?
Solution: $a = 100$, $d = -5$, $a_n = 5$. $5 = 100 - 5(n-1) \implies 5(n-1) = 95 \implies n-1 = 19 \implies n = 20$.
Question 9: Which term of the AP $2, 5, 8, \dots$ is $32$?
Solution: $a = 2$, $d = 3$, $a_n = 32$. $32 = 2 + (n-1)3 \implies 30 = 3(n-1) \implies n-1 = 10 \implies n = 11$.
Question 10: Which term of the AP $10, 7, 4, \dots$ is $-20$?
Solution: $a = 10$, $d = -3$, $a_n = -20$. $-20 = 10 - 3(n-1) \implies -30 = -3(n-1) \implies n-1 = 10 \implies n = 11$.
Question 11: Find the sum of the first $10$ terms of the AP: $2, 4, 6, 8, \dots$
Solution: $n = 10$, $a = 2$, $d = 2$. $S_n = \frac{n}{2} [2a + (n-1)d] \implies S_{10} = \frac{10}{2} [2(2) + 9(2)] = 5 [4 + 18] = 5 \times 22 = 110$.
Question 12: Find the sum of the first $20$ odd positive integers.
Solution: The sequence is $1, 3, 5, \dots$ Here $a=1, d=2, n=20$. The sum of the first $n$ odd integers is always $n^2$. Therefore, $S_{20} = 20^2 = 400$.
Question 13: Calculate the sum of the first $15$ terms of the AP: $5, 10, 15, \dots$
Solution: $n = 15$, $a = 5$, $d = 5$. $S_{15} = \frac{15}{2} [2(5) + 14(5)] = \frac{15}{2} [10 + 70] = \frac{15}{2} \times 80 = 15 \times 40 = 600$.
Question 14: Evaluate the sum: $2 + 4 + 6 + \dots + 40$
Solution: This is an AP with $a=2, d=2, l=40$. First find $n$: $40 = 2 + (n-1)2 \implies 2n = 40 \implies n = 20$. Then $S_{20} = \frac{n}{2}(a+l) = \frac{20}{2}(2 + 40) = 10(42) = 420$.
Question 15: Evaluate the sum: $100 + 90 + 80 + \dots + 10$
Solution: $a=100, d=-10, l=10$. Find $n$: $10 = 100 - 10(n-1) \implies 10(n-1) = 90 \implies n=10$. Sum $S_{10} = \frac{10}{2}(100 + 10) = 5(110) = 550$.
Question 16: The sum of the first $n$ terms of an AP is $S_n = 2n^2 + n$. Find the $n^{\text{th}}$ term.
Solution: $a_n = S_n - S_{n-1} = (2n^2 + n) - [2(n-1)^2 + (n-1)] = 2n^2 + n - (2n^2 - 4n + 2 + n - 1) = 4n - 1$.
Question 17: If the $n^{\text{th}}$ term of an AP is $a_n = 3n - 1$, find the sum of the first $5$ terms.
Solution: $a_1 = 3(1) - 1 = 2$. $a_5 = 3(5) - 1 = 14$. The sum $S_5 = \frac{5}{2}(a_1 + a_5) = \frac{5}{2}(2 + 14) = \frac{5}{2}(16) = 40$.
Question 18: Three numbers in AP have a sum of $15$ and a product of $105$. Find the numbers.
Solution: Let the numbers be $a-d, a, a+d$. Sum $= 3a = 15 \implies a = 5$. Product $= a(a^2 - d^2) = 105 \implies 5(25 - d^2) = 105 \implies 25 - d^2 = 21 \implies d^2 = 4 \implies d = \pm 2$. The numbers are $3, 5, 7$.
Question 19: Four numbers in AP sum to $20$, and the sum of their squares is $120$. Find the numbers.
Solution: Let the numbers be $a-3d, a-d, a+d, a+3d$. Sum $= 4a = 20 \implies a = 5$. Sum of squares $= (a-3d)^2 + (a-d)^2 + (a+d)^2 + (a+3d)^2 = 4a^2 + 20d^2 = 120 \implies 100 + 20d^2 = 120 \implies d^2 = 1 \implies d = \pm 1$. The numbers are $2, 4, 6, 8$.
Question 20: Insert $3$ arithmetic means between $2$ and $14$.
Solution: Let the sequence be $2, A_1, A_2, A_3, 14$. Here $a=2$ and $14$ is the $5^{\text{th}}$ term. $a_5 = a + 4d \implies 14 = 2 + 4d \implies 12 = 4d \implies d = 3$. The means are $5, 8, 11$.
Question 21: Insert $4$ arithmetic means between $10$ and $35$.
Solution: $a = 10$, $a_6 = 35 \implies 10 + 5d = 35 \implies 5d = 25 \implies d = 5$. The means are $15, 20, 25, 30$.
Question 22: If $a, b, c$ are in AP, prove that $2b = a + c$.
Solution: Since $a, b, c$ are in AP, the common difference between consecutive terms must be equal. Thus, $b - a = c - b$. Rearranging the terms yields $2b = a + c$.
Question 23: Find the value of $x$ if $x+1$, $3x$, and $4x+2$ are consecutive terms of an AP.
Solution: For the terms to be in AP, $2(3x) = (x+1) + (4x+2)$. Therefore, $6x = 5x + 3 \implies x = 3$.
Question 24: Find $x$ if $2x, 5x-1, 6x+2$ are in AP.
Solution: $2(5x-1) = 2x + (6x+2) \implies 10x - 2 = 8x + 2 \implies 2x = 4 \implies x = 2$.
Question 25: The $7^{\text{th}}$ term of an AP is $34$ and the $13^{\text{th}}$ term is $64$. Find the $10^{\text{th}}$ term.
Solution: Note that $10$ is the exact midpoint between $7$ and $13$. Therefore, $a_{10}$ is the arithmetic mean of $a_7$ and $a_{13}$. $a_{10} = \frac{34 + 64}{2} = \frac{98}{2} = 49$.
Question 26: If the sum of the first $7$ terms of an AP is $49$, and the sum of the first $17$ terms is $289$, find the sum of the first $n$ terms.
Solution: $S_7 = 49 \implies \frac{7}{2}(2a+6d) = 49 \implies a+3d = 7$. $S_{17} = 289 \implies \frac{17}{2}(2a+16d) = 289 \implies a+8d = 17$. Subtracting gives $5d = 10 \implies d = 2$. Then $a = 1$. The sum of first $n$ terms is $S_n = \frac{n}{2}(2(1) + (n-1)2) = n^2$.
Question 27: The ratio of the sum of $n$ terms of two APs is $(7n+1) : (4n+27)$. Find the ratio of their $11^{\text{th}}$ terms.
Solution: To find the ratio of the $k^{\text{th}}$ terms when the ratio of the sum of $n$ terms is given, substitute $n = 2k - 1$. Here $k = 11$, so substitute $n = 21$. Ratio $= \frac{7(21) + 1}{4(21) + 27} = \frac{147 + 1}{84 + 27} = \frac{148}{111} = \frac{4}{3}$.
Question 28: In an AP, if the $p^{\text{th}}$ term is $q$ and the $q^{\text{th}}$ term is $p$, find the $n^{\text{th}}$ term.
Solution: $a + (p-1)d = q$ and $a + (q-1)d = p$. Subtracting gives $(p-q)d = q-p \implies d = -1$. Substituting $d$, $a - p + 1 = q \implies a = p+q-1$. The $n^{\text{th}}$ term is $a_n = (p+q-1) + (n-1)(-1) = p+q-n$.
Question 29: How many multiples of $4$ lie between $10$ and $250$?
Solution: The first multiple after $10$ is $12$. The last multiple before $250$ is $248$. The AP is $12, 16, \dots, 248$. $a = 12$, $d = 4$, $a_n = 248$. $248 = 12 + (n-1)4 \implies 236 = 4(n-1) \implies n-1 = 59 \implies n = 60$.
Question 30: Find the sum of all two-digit odd numbers.
Solution: The sequence is $11, 13, 15, \dots, 99$. Here $a=11, d=2, l=99$. Number of terms: $99 = 11 + (n-1)2 \implies 88 = 2(n-1) \implies n=45$. Sum $= \frac{45}{2}(11 + 99) = \frac{45}{2}(110) = 45 \times 55 = 2475$.
Practice Questions: Geometric Progression (GP)
Question 31: Find the $5^{\text{th}}$ term of the GP: $2, 4, 8, \dots$
Solution: $a = 2$, common ratio $r = 4/2 = 2$. The $n^{\text{th}}$ term is $a_n = ar^{n-1}$. Thus, $a_5 = 2(2^4) = 2(16) = 32$.
Question 32: Find the $6^{\text{th}}$ term of the sequence: $3, 9, 27, \dots$
Solution: $a = 3$, $r = 3$. $a_6 = 3(3^5) = 3(243) = 729$.
Question 33: Calculate the $7^{\text{th}}$ term of the GP: $1, \frac{1}{2}, \frac{1}{4}, \dots$
Solution: $a = 1$, $r = \frac{1}{2}$. $a_7 = 1\left(\frac{1}{2}\right)^6 = \frac{1}{64}$.
Question 34: In a GP, the first term is $2$ and the $4^{\text{th}}$ term is $54$. Find the common ratio.
Solution: $a = 2$, $a_4 = 54$. Since $a_4 = ar^3$, we have $2r^3 = 54 \implies r^3 = 27 \implies r = 3$.
Question 35: If the $2^{\text{nd}}$ term of a GP is $10$ and the $5^{\text{th}}$ term is $1250$, find the first term $a$.
Solution: $ar = 10$ and $ar^4 = 1250$. Dividing the second equation by the first: $r^3 = 125 \implies r = 5$. Substitute $r$ back: $a(5) = 10 \implies a = 2$.
Question 36: How many terms are in the GP: $2, 4, 8, \dots, 256$?
Solution: $a = 2$, $r = 2$, $a_n = 256$. Using $ar^{n-1} = 256 \implies 2(2^{n-1}) = 256 \implies 2^n = 256 \implies n = 8$.
Question 37: Find the number of terms in the GP: $3, 6, 12, \dots, 384$.
Solution: $a = 3$, $r = 2$, $a_n = 384$. $3(2^{n-1}) = 384 \implies 2^{n-1} = 128 = 2^7 \implies n-1 = 7 \implies n = 8$.
Question 38: Which term of the GP $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots$ is $\frac{1}{2187}$?
Solution: $a = \frac{1}{3}$, $r = \frac{1}{3}$. $a_n = \frac{1}{3}\left(\frac{1}{3}\right)^{n-1} = \left(\frac{1}{3}\right)^n$. So, $\left(\frac{1}{3}\right)^n = \frac{1}{2187} = \frac{1}{3^7} \implies n = 7$.
Question 39: Which term of the progression $\sqrt{2}, 2, 2\sqrt{2}, \dots$ is $16$?
Solution: $a = \sqrt{2}$, $r = \frac{2}{\sqrt{2}} = \sqrt{2}$. $a_n = \sqrt{2}(\sqrt{2})^{n-1} = (\sqrt{2})^n = 16$. Since $16 = 2^4 = (\sqrt{2})^8$, it implies $n = 8$.
Question 40: The $3^{\text{rd}}$ term of a GP is $24$ and the $6^{\text{th}}$ term is $192$. Find the $10^{\text{th}}$ term.
Solution: $ar^2 = 24$ and $ar^5 = 192$. Dividing gives $r^3 = 8 \implies r = 2$. Then $a(4) = 24 \implies a = 6$. The $10^{\text{th}}$ term is $a_{10} = 6(2^9) = 6(512) = 3072$.
Question 41: Find the sum of the first $5$ terms of the GP: $2, 6, 18, \dots$
Solution: $a = 2$, $r = 3$. The sum $S_n = \frac{a(r^n - 1)}{r - 1}$. $S_5 = \frac{2(3^5 - 1)}{3 - 1} = \frac{2(243 - 1)}{2} = 242$.
Question 42: Find the sum of the first $6$ terms of the sequence: $1, 2, 4, 8, \dots$
Solution: $a = 1$, $r = 2$. $S_6 = \frac{1(2^6 - 1)}{2 - 1} = 64 - 1 = 63$.
Question 43: Calculate the sum of the first $4$ terms of the GP: $5, -10, 20, \dots$
Solution: $a = 5$, $r = -2$. $S_4 = \frac{5(1 - (-2)^4)}{1 - (-2)} = \frac{5(1 - 16)}{3} = \frac{5(-15)}{3} = -25$.
Question 44: Find the sum to infinity of the GP: $1, \frac{1}{2}, \frac{1}{4}, \dots$
Solution: $a = 1$, $r = \frac{1}{2}$. Since $|r| < 1$, the sum to infinity is $S_\infty = \frac{a}{1 - r} = \frac{1}{1 - 1/2} = \frac{1}{1/2} = 2$.
Question 45: Evaluate the sum of the infinite series: $5 + 1 + \frac{1}{5} + \dots$
Solution: $a = 5$, $r = \frac{1}{5}$. $S_\infty = \frac{5}{1 - 1/5} = \frac{5}{4/5} = \frac{25}{4}$.
Question 46: Find the sum to infinity of $10, -5, \frac{5}{2}, \dots$
Solution: $a = 10$, $r = -\frac{1}{2}$. $S_\infty = \frac{10}{1 - (-1/2)} = \frac{10}{3/2} = \frac{20}{3}$.
Question 47: Find the value of $x$ if $x$, $x+2$, and $x+6$ are consecutive terms of a GP.
Solution: For terms to be in GP, the square of the middle term equals the product of the extremes: $(x+2)^2 = x(x+6) \implies x^2 + 4x + 4 = x^2 + 6x \implies 4 = 2x \implies x = 2$.
Question 48: Find $x$ if $x-1$, $2x$, and $5x+3$ are in GP.
Solution: $(2x)^2 = (x-1)(5x+3) \implies 4x^2 = 5x^2 - 2x - 3 \implies x^2 - 2x - 3 = 0$. Factoring gives $(x-3)(x+1) = 0 \implies x = 3$ or $x = -1$.
Question 49: Insert $2$ geometric means between $3$ and $24$.
Solution: Let the GP be $3, G_1, G_2, 24$. Then $24$ is the $4^{\text{th}}$ term. $ar^3 = 24 \implies 3r^3 = 24 \implies r^3 = 8 \implies r = 2$. The means are $6, 12$.
Question 50: Insert $3$ geometric means between $2$ and $162$.
Solution: $a = 2$, $a_5 = 162$. $ar^4 = 162 \implies 2r^4 = 162 \implies r^4 = 81$. Assuming real and positive terms, $r = 3$. The means are $6, 18, 54$.
Question 51: The $3^{\text{rd}}$ term of a GP is $4$. Find the product of its first $5$ terms.
Solution: Let the $5$ terms be $\frac{a}{r^2}, \frac{a}{r}, a, ar, ar^2$. Their product is $a^5$. The $3^{\text{rd}}$ term is $a$, so $a = 4$. Product $= 4^5 = 1024$.
Question 52: The sum of the first two terms of a GP is $12$ and the $5^{\text{th}}$ term is $4$ times the $3^{\text{rd}}$ term. Find the GP assuming the common ratio is positive.
Solution: $ar^4 = 4ar^2 \implies r^2 = 4 \implies r = 2$ (since $r > 0$). Given $a + ar = 12 \implies a + 2a = 12 \implies 3a = 12 \implies a = 4$. The GP is $4, 8, 16, \dots$
Question 53: If $a, b, c$ are in GP, prove that $b^2 = ac$.
Solution: For terms to be in GP, the ratio of consecutive terms must be equal: $\frac{b}{a} = \frac{c}{b}$. Cross-multiplying gives $b^2 = ac$.
Question 54: The sum of an infinite GP is $15$, and the sum of the squares of its terms is $45$. Find the common ratio $r$.
Solution: $\frac{a}{1-r} = 15$ and $\frac{a^2}{1-r^2} = 45$. Divide the second by the square of the first: $\frac{a^2 / (1-r)(1+r)}{a^2 / (1-r)^2} = \frac{45}{225} \implies \frac{1-r}{1+r} = \frac{1}{5} \implies 5 - 5r = 1 + r \implies 6r = 4 \implies r = \frac{2}{3}$.
Question 55: A GP consists of an even number of terms. If the sum of all terms is $3$ times the sum of the odd-positioned terms, find the common ratio.
Solution: $S_{\text{total}} = S_{\text{odd}} + S_{\text{even}}$. Since each even term is $r$ times the previous odd term, $S_{\text{even}} = r \times S_{\text{odd}}$. Total sum $= (1+r)S_{\text{odd}}$. Given $(1+r)S_{\text{odd}} = 3S_{\text{odd}} \implies 1+r = 3 \implies r = 2$.
Question 56: Express the repeating decimal $0.\overline{3}$ as a fraction using infinite GP.
Solution: $0.\overline{3} = \frac{3}{10} + \frac{3}{100} + \dots$ This is a GP with $a = \frac{3}{10}, r = \frac{1}{10}$. $S_\infty = \frac{3/10}{1 - 1/10} = \frac{3/10}{9/10} = \frac{1}{3}$.
Question 57: Express the repeating decimal $0.\overline{45}$ as a fraction using GP.
Solution: $0.\overline{45} = \frac{45}{100} + \frac{45}{10000} + \dots$ Here $a = \frac{45}{100}$ and $r = \frac{1}{100}$. $S_\infty = \frac{45/100}{1 - 1/100} = \frac{45}{99} = \frac{5}{11}$.
Question 58: Find the sum of $n$ terms of the sequence: $1, 11, 111, \dots$
Solution: $S_n = 1 + 11 + 111 + \dots = \frac{1}{9}(9 + 99 + 999 + \dots) = \frac{1}{9}[(10-1) + (10^2-1) + (10^3-1) + \dots] = \frac{1}{9}\left[ \frac{10(10^n-1)}{9} - n \right]$.
Question 59: The first term of a GP is $1$. The sum of the $3^{\text{rd}}$ and $5^{\text{th}}$ terms is $90$. Find the positive common ratio.
Solution: $a = 1$. $ar^2 + ar^4 = 90 \implies r^2 + r^4 - 90 = 0$. Let $x = r^2$, then $x^2 + x - 90 = 0 \implies (x+10)(x-9) = 0$. Since $x = r^2 \geq 0$, $r^2 = 9 \implies r = 3$.
Question 60: Find the minimum value of integer $n$ for which the sum of the first $n$ terms of $1, 3, 9, 27, \dots$ exceeds $1000$.
Solution: $a = 1, r = 3$. $S_n = \frac{1(3^n - 1)}{3 - 1} = \frac{3^n - 1}{2}$. We need $\frac{3^n - 1}{2} > 1000 \implies 3^n - 1 > 2000 \implies 3^n > 2001$. Since $3^6 = 729$ and $3^7 = 2187$, the minimum value is $n = 7$.