CBSE & State Board Question Bank | 9th - 12th

Deepa Maths Academy • Global Examination Portal

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

Secure University-Grade Repository for Model Assessments, Board Examinations, and Step-by-Step Solutions.

Portal ID: DMA-EXAM-2026 • Security: Encrypted • Status: Active Examination

SECTION E — Long Answer Type Questions [5 Marks Each]

• Q1. A manufacturer of TV sets produced 600 sets in the 3rd year and 700 sets in the 7th year. Assuming that the production increases uniformly by a fixed number every year, find:
  (i) the production in the 1st year,
  (ii) the production in the 10th year,
  (iii) the total production in the first 7 years.
  Answer:
  1. Let the production in the 1st year be $a$ and the fixed annual increase (common difference) be $d$.
  
  2. We are given the production in the 3rd year ($a_3 = 600$) and the 7th year ($a_7 = 700$).
      $a_3 = a + 2d = 600$
      $a_7 = a + 6d = 700$
  
  3. Subtracting the first equation from the second equation:
      $(a + 6d) - (a + 2d) = 700 - 600$
      $4d = 100 \implies d = 25$
  
  4. Substituting $d = 25$ into the first equation ($a + 2(25) = 600$):
      $a + 50 = 600 \implies a = 550$
  
  5. (i) Production in the 1st year ($a$): $550$ TV sets.
  
  6. (ii) Production in the 10th year ($a_{10}$): Using $a_n = a + (n - 1)d$:
      $a_{10} = 550 + (10 - 1) \times 25$
      $a_{10} = 550 + 9 \times 25$
      $a_{10} = 550 + 225 = 775$ TV sets.
  
  7. (iii) Total production in the first 7 years ($S_7$): Using $S_n = \frac{n}{2}[2a + (n - 1)d]$:
      $S_7 = \frac{7}{2}[2(550) + (7 - 1) \times 25]$
      $S_7 = \frac{7}{2}[1100 + 6 \times 25]$
      $S_7 = \frac{7}{2}[1100 + 150]$
      $S_7 = \frac{7}{2} \times 1250 = 7 \times 625 = 4375$ TV sets.
  
  Result:
  (i) Production in the 1st year = 550
  (ii) Production in the 10th year = 775
  (iii) Total production in the first 7 years = 4375
• Q2. Subba Rao started work in 1995 at an annual salary of ₹5,000 and received an increment of ₹200 each year. In which year did his income reach ₹7,000?
  Answer:
  1. The annual salaries form an Arithmetic Progression (AP) where:
      * First term ($a$) = ₹5,000
      * Common difference ($d$) = ₹200
      * Let his salary in the $n$-th year be $a_n = \text{₹}7,000$
  
  2. Using the formula for the $n$-th term of an AP ($a_n = a + (n - 1)d$):
      $7000 = 5000 + (n - 1) \times 200$
  
  3. Solving for $n$:
      $7000 - 5000 = (n - 1) \times 200$
      $2000 = (n - 1) \times 200$
      $n - 1 = \frac{2000}{200} = 10$
      $n = 10 + 1 = 11$
  
  4. Calculating the corresponding calendar year:
      Starting year = 1995
      Calendar Year = $1995 + (n - 1) = 1995 + 10 = 2005$
  
  Result: Therefore, Subba Rao's income reached ₹7,000 in the 11th year, which corresponds to the year **2005**.
• Q3. Ramkali saved ₹5 in the first week of a year and then increased her weekly savings by ₹1.75. If in the n^th week, her weekly savings become ₹20.75, find n.
  Answer:
  1. The weekly savings form an Arithmetic Progression (AP) where:
      * First term ($a$) = ₹5.00
      * Common difference ($d$) = ₹1.75
      * $n$-th term ($a_n$) = ₹20.75
  
  2. Using the formula for the $n$-th term of an AP ($a_n = a + (n - 1)d$):
      $20.75 = 5.00 + (n - 1) \times 1.75$
  
  3. Solving for $n$:
      $20.75 - 5.00 = (n - 1) \times 1.75$
      $15.75 = (n - 1) \times 1.75$
      $n - 1 = \frac{15.75}{1.75} = 9$
      $n = 9 + 1 = 10$
  
  Result: Therefore, the value of $n$ is 10.
• Q4. A sum of ₹1,000 is invested at $8\%$ simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30 years using this fact.
  Answer:
  1. The formula for simple interest is given by $\text{SI} = \frac{P \times R \times T}{100}$, where Principal ($P$) = ₹1,000 and Rate ($R$) = $8\%$.
  
  2. Calculating the interest at the end of each year ($T$):
      * Interest at the end of 1st year ($I_1$) = $\frac{1000 \times 8 \times 1}{100} = \text{₹}80$
      * Interest at the end of 2nd year ($I_2$) = $\frac{1000 \times 8 \times 2}{100} = \text{₹}160$
      * Interest at the end of 3rd year ($I_3$) = $\frac{1000 \times 8 \times 3}{100} = \text{₹}240$
  
  3. Checking if these interests form an AP:
      The sequence of interests is ₹80, ₹160, ₹240, $\dots$
      Common difference ($d$) = $I_2 - I_1 = 160 - 80 = 80$, and $I_3 - I_2 = 240 - 160 = 80$.
      Since the difference between consecutive terms is constant, yes, these interests form an AP with first term ($a$) = $80$ and common difference ($d$) = $80$.
  
  4. Finding the interest at the end of 30 years ($a_{30}$) using the AP formula $a_n = a + (n - 1)d$:
      $a_{30} = 80 + (30 - 1) \times 80$
      $a_{30} = 80 + 29 \times 80$
      $a_{30} = 80 + 2320 = 2400$
  
  Result: Therefore, the interest at the end of 30 years is ₹2,400.
• Q5. In a potato race, a bucket is placed at the starting point, which is $5\text{ m}$ from the first potato, and the other potatoes are placed $3\text{ m}$ apart in a straight line. There are ten potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
  Answer:
  1. Distance of the first potato from the bucket = $5\text{ m}$.
      Total distance run to retrieve the 1st potato ($a_1$) = $2 \times 5 = 10\text{ m}$.
  
  2. Distance of the second potato from the bucket = $5 + 3 = 8\text{ m}$.
      Total distance run to retrieve the 2nd potato ($a_2$) = $2 \times 8 = 16\text{ m}$.
  
  3. Distance of the third potato from the bucket = $5 + 3 + 3 = 11\text{ m}$.
      Total distance run to retrieve the 3rd potato ($a_3$) = $2 \times 11 = 22\text{ m}$.
  
  4. The total distances run for each potato form an AP: 10, 16, 22, $\dots$
      Here, first term ($a$) = 10, common difference ($d$) = $16 - 10 = 6$, and number of terms ($n$) = 10.
  
  5. Using the AP sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$ to find the total distance for 10 potatoes ($S_{10}$):
      $S_{10} = \frac{10}{2}[2(10) + (10 - 1) \times 6]$
  
  6. Simplifying the expression:
      $S_{10} = 5[20 + 9 \times 6]$
      $S_{10} = 5[20 + 54]$
      $S_{10} = 5 \times 74 = 370$
  
  Result: Therefore, the total distance the competitor has to run is $370\text{ m}$.
• Q6. 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 the bottom rungs are $2\frac{1}{2}\text{ m}$ apart, what is the length of the wood required for the rungs?
  Answer:
  1. Length of the bottom rung ($a$) = $45\text{ cm}$, length of the top rung ($l$ or $a_n$) = $25\text{ cm}$.
  
  2. Distance between any two consecutive rungs = $25\text{ cm}$.
  
  3. Total distance between the top and bottom rungs = $2\frac{1}{2}\text{ m} = 2.5\text{ m} = 250\text{ cm}$.
  
  4. Finding the total number of rungs ($n$):
      $n = \frac{\text{Total distance}}{\text{Distance between rungs}} + 1$
      $n = \frac{250}{25} + 1 = 10 + 1 = 11$
  
  5. The lengths of the rungs form an Arithmetic Progression (AP) with $n = 11$, first term $a = 45$, and last term $a_{11} = 25$.
  
  6. Using the AP sum formula $S_n = \frac{n}{2}(a + a_n)$ to find the total length of wood required ($S_{11}$):
      $S_{11} = \frac{11}{2}(45 + 25)$
      $S_{11} = \frac{11}{2} \times 70$
      $S_{11} = 11 \times 35 = 385$
  
  Result: Therefore, the total length of the wood required for the rungs is $385\text{ cm}$ (or $3.85\text{ m}$).
• Q7. Houses in a row are numbered consecutively from 1 to 49. Show that there is a value of $x$ such that the sum of the numbers of the houses preceding the house numbered $x$ is equal to the sum of the numbers of the houses following it. Find this value of $x$.
  Answer:
  1. The house numbers form an AP: $1, 2, 3, \dots, 49$, where $a = 1$ and $d = 1$.
  
  2. Let the house numbered $x$ be somewhere in the middle.
      * The sum of the house numbers preceding house $x$ is the sum of numbers from 1 to $(x - 1)$, denoted as $S_{x-1}$.
      * The sum of the house numbers following house $x$ is the sum of numbers from $(x + 1)$ to 49, which can be written as $S_{49} - S_x$.
  
  3. According to the given condition, these two sums are equal:
      $S_{x-1} = S_{49} - S_x$
  
  4. Using the sum formula $S_n = \frac{n(n + 1)}{2}$:
      * $S_{x-1} = \frac{(x - 1)[(x - 1) + 1]}{2} = \frac{(x - 1)x}{2}$
      * $S_x = \frac{x(x + 1)}{2}$
      * $S_{49} = \frac{49(49 + 1)}{2} = \frac{49 \times 50}{2} = 49 \times 25 = 1225$
  
  5. Substituting these into the equation:
      $\frac{x(x - 1)}{2} = 1225 - \frac{x(x + 1)}{2}$
  
  6. Rearranging the terms:
      $\frac{x(x + 1)}{2} + \frac{x(x - 1)}{2} = 1225$
      $\frac{x^2 + x + x^2 - x}{2} = 1225$
      $\frac{2x^2}{2} = 1225$
      $x^2 = 1225$
  
  7. Solving for $x$ (since house numbers must be positive):
      $x = \sqrt{1225} = 35$
  
  Result: Therefore, the value of $x$ is 35.
• Q8. If the $m^{\text{th}}$ term of an AP is $\frac{1}{n}$ and the $n^{\text{th}}$ term is $\frac{1}{m}$, then show that its $(mn)^{\text{th}}$ term is $1$.
  Answer:
  1. Let the first term of the AP be $a$ and the common difference be $d$.
  
  2. Using the formula for the $k$-th term of an AP ($a_k = a + (k - 1)d$), we are given:
      * $a_m = a + (m - 1)d = \frac{1}{n}$     --- (Equation 1)
      * $a_n = a + (n - 1)d = \frac{1}{m}$     --- (Equation 2)
  
  3. Subtracting Equation 2 from Equation 1 to find $d$:
      $[(m - 1) - (n - 1)]d = \frac{1}{n} - \frac{1}{m}$
      $(m - n)d = \frac{m - n}{mn}$
      $d = \frac{1}{mn}$
  
  4. Substituting the value of $d$ into Equation 1 to find $a$:
      $a + (m - 1)\left(\frac{1}{mn}\right) = \frac{1}{n}$
      $a + \frac{m - 1}{mn} = \frac{1}{n}$
      $a = \frac{1}{n} - \frac{m - 1}{mn}$
      $a = \frac{m - (m - 1)}{mn} = \frac{m - m + 1}{mn} = \frac{1}{mn}$
  
  5. Now, we need to find the $(mn)^{\text{th}}$ term ($a_{mn}$):
      $a_{mn} = a + (mn - 1)d$
  
  6. Substituting the values of $a$ and $d$:
      $a_{mn} = \frac{1}{mn} + (mn - 1)\left(\frac{1}{mn}\right)$
      $a_{mn} = \frac{1 + mn - 1}{mn}$
      $a_{mn} = \frac{mn}{mn} = 1$
  
  Result: Hence, it is proved that the $(mn)^{\text{th}}$ term of the AP is $1$.
• Q9. If the sum of the first $p$ terms of an AP is equal to the sum of the first $q$ terms, then show that the sum of its first $(p+q)$ terms is zero.
  Answer:
  1. Let the first term of the AP be $a$ and the common difference be $d$.
  
  2. The formula for the sum of the first $n$ terms of an AP is $S_n = \frac{n}{2}[2a + (n - 1)d]$.
  
  3. Given that the sum of the first $p$ terms is equal to the sum of the first $q$ terms ($S_p = S_q$):
      $\frac{p}{2}[2a + (p - 1)d] = \frac{q}{2}[2a + (q - 1)d]$
  
  4. Multiplying both sides by 2 and expanding the terms:
      $p[2a + (p - 1)d] = q[2a + (q - 1)d]$
      $2ap + p(p - 1)d = 2aq + q(q - 1)d$
  
  5. Rearranging all terms to one side:
      $2ap - 2aq + p(p - 1)d - q(q - 1)d = 0$
      $2a(p - q) + [p^2 - p - (q^2 - q)]d = 0$
      $2a(p - q) + [(p^2 - q^2) - (p - q)]d = 0$
  
  6. Factoring $(p^2 - q^2)$ as $(p - q)(p + q)$:
      $2a(p - q) + [(p - q)(p + q) - (p - q)]d = 0$
      $2a(p - q) + (p - q)(p + q - 1)d = 0$
  
  7. Taking $(p - q)$ common out (assuming $p \neq q$):
      $(p - q)[2a + (p + q - 1)d] = 0$
      Since $p \neq q$, we get:
      $2a + (p + q - 1)d = 0$     --- (Equation 1)
  
  8. Now, we need to find the sum of the first $(p + q)$ terms ($S_{p+q}$):
      $S_{p+q} = \frac{p + q}{2}[2a + (p + q - 1)d]$
  
  9. Substituting Equation 1 into the sum formula:
      $S_{p+q} = \frac{p + q}{2}[0] = 0$
  
  Result: Hence, it is proved that the sum of its first $(p + q)$ terms is zero.
• Q10. If the sum of first $n$ terms of an AP is given by $S_n = 3n^2 + 5n$, find the $n^{\text{th}}$ term of this AP and hence find its 16th term.
  Answer:
  1. We are given the sum of the first $n$ terms of the AP as $S_n = 3n^2 + 5n$.
  
  2. The $n^{\text{th}}$ term of an AP can be found using the relation $a_n = S_n - S_{n-1}$.
  
  3. First, find $S_{n-1}$ by replacing $n$ with $(n - 1)$ in $S_n$:
      $S_{n-1} = 3(n - 1)^2 + 5(n - 1)$
      $S_{n-1} = 3(n^2 - 2n + 1) + 5n - 5$
      $S_{n-1} = 3n^2 - 6n + 3 + 5n - 5$
      $S_{n-1} = 3n^2 - n - 2$
  
  4. Now, calculate the $n^{\text{th}}$ term ($a_n$):
      $a_n = S_n - S_{n-1}$
      $a_n = (3n^2 + 5n) - (3n^2 - n - 2)$
      $a_n = 3n^2 + 5n - 3n^2 + n + 2$
      $a_n = 6n + 2$
  
  5. Next, find the 16th term ($a_{16}$) by substituting $n = 16$ into the $n^{\text{th}}$ term formula:
      $a_{16} = 6(16) + 2$
      $a_{16} = 96 + 2 = 98$
  
  Result: Therefore, the $n^{\text{th}}$ term of the AP is $6n + 2$ and its 16th term is $98$.
• Q11. If the ratio of the sum of the first $m$ and $n$ terms of an AP is $m^2 : n^2$, show that the ratio of its $m^{\text{th}}$ and $n^{\text{th}}$ terms is $(2m - 1) : (2n - 1)$.
  Answer:
  1. Let the first term of the AP be $a$ and the common difference be $d$.
  
  2. The formula for the sum of the first $k$ terms is $S_k = \frac{k}{2}[2a + (k - 1)d]$.
  
  3. We are given the ratio of the sum of the first $m$ terms to the sum of the first $n$ terms:
      $\frac{S_m}{S_n} = \frac{m^2}{n^2}$
  
  4. Substituting the sum formula for $m$ and $n$ terms:
      $\frac{\frac{m}{2}[2a + (m - 1)d]}{\frac{n}{2}[2a + (n - 1)d]} = \frac{m^2}{n^2}$
  
  5. Simplifying the expression by cancelling $\frac{1}{2}$ and one $m$ and $n$ from both numerator and denominator:
      $\frac{2a + (m - 1)d}{2a + (n - 1)d} = \frac{m}{n}$
  
  6. Cross-multiplying to solve further:
      $n[2a + (m - 1)d] = m[2a + (n - 1)d]$
      $2an + n(m - 1)d = 2am + m(n - 1)d$
  
  7. Rearranging the terms containing $2a$ on one side and $d$ on the other:
      $2an - 2am = m(n - 1)d - n(m - 1)d$
      $2a(n - m) = [mn - m - mn + n]d$
      $2a(n - m) = (n - m)d$
  
  8. Dividing both sides by $(n - m)$ (assuming $m \neq n$):
      $2a = d$
  
  9. Now, we need to find the ratio of the $m^{\text{th}}$ and $n^{\text{th}}$ terms ($a_m : a_n$):
      $\frac{a_m}{a_n} = \frac{a + (m - 1)d}{a + (n - 1)d}$
  
  10. Substituting $d = 2a$ into the ratio:
      $\frac{a_m}{a_n} = \frac{a + (m - 1)(2a)}{a + (n - 1)(2a)}$
  
  11. Taking $a$ common out from numerator and denominator and cancelling it:
      $\frac{a_m}{a_n} = \frac{1 + 2(m - 1)}{1 + 2(n - 1)}$
      $\frac{a_m}{a_n} = \frac{1 + 2m - 2}{1 + 2n - 2}$
      $\frac{a_m}{a_n} = \frac{2m - 1}{2n - 1}$
  
  Result: Hence, it is proved that the ratio of its $m^{\text{th}}$ and $n^{\text{th}}$ terms is $(2m - 1) : (2n - 1)$.
• Q12. The sum of the $4^{\text{th}}$ and $8^{\text{th}}$ terms of an AP is 24 and the sum of the $6^{\text{th}}$ and $10^{\text{th}}$ terms is 44. Find the first term and the common difference of the AP.
  Answer:
  1. Let the first term of the AP be $a$ and the common difference be $d$.
  
  2. Using the formula for the $k$-th term ($a_k = a + (k - 1)d$), we can write expressions for the given terms:
      * $a_4 = a + 3d$
      * $a_8 = a + 7d$
      * $a_6 = a + 5d$
      * $a_{10} = a + 9d$
  
  3. From the first given condition, the sum of the $4^{\text{th}}$ and $8^{\text{th}}$ terms is 24:
      $(a + 3d) + (a + 7d) = 24$
      $2a + 10d = 24$
      Dividing by 2:
      $a + 5d = 12$     --- (Equation 1)
  
  4. From the second given condition, the sum of the $6^{\text{th}}$ and $10^{\text{th}}$ terms is 44:
      $(a + 5d) + (a + 9d) = 44$
      $2a + 14d = 44$
      Dividing by 2:
      $a + 7d = 22$     --- (Equation 2)
  
  5. Subtracting Equation 1 from Equation 2 to find $d$:
      $(a + 7d) - (a + 5d) = 22 - 12$
      $2d = 10 \implies d = 5$
  
  6. Substituting the value of $d = 5$ into Equation 1 to find $a$:
      $a + 5(5) = 12$
      $a + 25 = 12$
      $a = 12 - 25 = -13$
  
  Result: Therefore, the first term ($a$) of the AP is $-13$ and the common difference ($d$) is $5$.
• Q13. If the $p^{\text{th}}$, $q^{\text{th}}$ and $r^{\text{th}}$ terms of an AP are $a, b$ and $c$ respectively, then show that: $$a(q - r) + b(r - p) + c(p - q) = 0$$
  Answer:
  1. Let the first term of the given AP be $A$ and the common difference be $D$ (using capital letters to avoid confusion with the terms $a, b, c$).
  
  2. Using the formula for the $k^{\text{th}}$ term of an AP ($a_k = A + (k - 1)D$), we are given:
      * $A_p = A + (p - 1)D = a$     --- (Equation 1)
      * $A_q = A + (q - 1)D = b$     --- (Equation 2)
      * $A_r = A + (r - 1)D = c$     --- (Equation 3)
  
  3. We need to prove that $a(q - r) + b(r - p) + c(p - q) = 0$.
  
  4. Substituting the values of $a$, $b$, and $c$ from Equations 1, 2, and 3 into the Left-Hand Side (LHS):
      $\text{LHS} = [A + (p - 1)D](q - r) + [A + (q - 1)D](r - p) + [A + (r - 1)D](p - q)$
  
  5. Grouping the terms containing $A$ and the terms containing $D$ separately:
      $\text{LHS} = A[(q - r) + (r - p) + (p - q)] + D[(p - 1)(q - r) + (q - 1)(r - p) + (r - 1)(p - q)]$
  
  6. Simplifying the term inside the first bracket (for $A$):
      $(q - r + r - p + p - q) = 0$
      So, the first part becomes $A \times 0 = 0$.
  
  7. Expanding the products inside the second bracket (for $D$):
      $= (pq - pr - q + r) + (qr - pq - r + p) + (rp - rq - p + q)$
  
  8. Combining all the terms inside the second bracket:
      $= pq - pr - q + r + qr - pq - r + p + rp - rq - p + q$
      Notice that all terms cancel each other out ($pq - pq$, $-pr + rp$, $-q + q$, $+r - r$, $+qr - rq$, $-p + p$):
      $= 0$
  
  9. Therefore, the expression simplifies to:
      $\text{LHS} = A(0) + D(0) = 0 + 0 = 0 = \text{RHS}$
  
  Result: Hence, it is proved that $a(q - r) + b(r - p) + c(p - q) = 0$.
• Q14. An AP consists of 50 terms of which the $3^{\text{rd}}$ term is 12 and the last term is 106. Find the $29^{\text{th}}$ term.
  Answer:
  1. Let the first term of the AP be $a$, the common difference be $d$, and the total number of terms be $n = 50$.
  
  2. We are given the $3^{\text{rd}}$ term ($a_3 = 12$) and the last term (which is the $50^{\text{th}}$ term, $a_{50} = 106$).
      * $a_3 = a + 2d = 12$     --- (Equation 1)
      * $a_{50} = a + 49d = 106$     --- (Equation 2)
  
  3. Subtracting Equation 1 from Equation 2 to find $d$:
      $(a + 49d) - (a + 2d) = 106 - 12$
      $47d = 94$
      $d = \frac{94}{47} = 2$
  
  4. Substituting the value of $d = 2$ into Equation 1 to find $a$:
      $a + 2(2) = 12$
      $a + 4 = 12$
      $a = 12 - 4 = 8$
  
  5. Now, we need to find the $29^{\text{th}}$ term ($a_{29}$) using the formula $a_n = a + (n - 1)d$:
      $a_{29} = 8 + (29 - 1) \times 2$
      $a_{29} = 8 + 28 \times 2$
      $a_{29} = 8 + 56 = 64$
  
  Result: Therefore, the $29^{\text{th}}$ term of the AP is $64$.
• Q15. If the ratio of the sum of first $n$ terms of two APs is $(7n + 1) : (4n + 27)$, find the ratio of their $9^{\text{th}}$ terms.
  Answer:
  1. Let the first terms of the two APs be $a_1$ and $a_2$, and their common differences be $d_1$ and $d_2$ respectively.
  
  2. Let $S_{n1}$ and $S_{n2}$ be the sums of the first $n$ terms of the two APs. We are given the ratio of their sums:
      $\frac{S_{n1}}{S_{n2}} = \frac{7n + 1}{4n + 27}$
  
  3. Using the formula for the sum of an AP ($S_n = \frac{n}{2}[2a + (n - 1)d]$):
      $\frac{\frac{n}{2}[2a_1 + (n - 1)d_1]}{\frac{n}{2}[2a_2 + (n - 1)d_2]} = \frac{7n + 1}{4n + 27}$
      $\frac{2a_1 + (n - 1)d_1}{2a_2 + (n - 1)d_2} = \frac{7n + 1}{4n + 27}$     --- (Equation 1)
  
  4. We need to find the ratio of their $9^{\text{th}}$ terms ($a_{9, 1} : a_{9, 2}$):
      $\frac{a_{9, 1}}{a_{9, 2}} = \frac{a_1 + 8d_1}{a_2 + 8d_2}$
  
  5. Multiply both the numerator and the denominator of the Left-Hand Side of Equation 1 by 2 so that it matches the form of the $9^{\text{th}}$ term:
      $\frac{a_1 + \frac{n - 1}{2} d_1}{a_2 + \frac{n - 1}{2} d_2} = \frac{7n + 1}{4n + 27}$
  
  6. To find the ratio of the $9^{\text{th}}$ terms, we equate the coefficient of $d$ to 8:
      $\frac{n - 1}{2} = 8 \implies n - 1 = 16 \implies n = 17$
  
  7. Substituting $n = 17$ into Equation 1:
      $\frac{a_1 + 8d_1}{a_2 + 8d_2} = \frac{7(17) + 1}{4(17) + 27}$
      $\frac{a_1 + 8d_1}{a_2 + 8d_2} = \frac{119 + 1}{68 + 27}$
      $\frac{a_1 + 8d_1}{a_2 + 8d_2} = \frac{120}{95}$
  
  8. Simplifying the fraction by dividing numerator and denominator by 5:
      $\frac{120}{95} = \frac{24}{19}$
  
  Result: Therefore, the ratio of their $9^{\text{th}}$ terms is $24 : 19$.
• Q16. The sum of three numbers in an AP is 21 and the sum of their squares is 165. Find the numbers.
  Answer:
  1. When three numbers are in an AP, it is convenient to choose them as $(a - d)$, $a$, and $(a + d)$, where $a$ is the middle term and $d$ is the common difference.
  
  2. From the first given condition, the sum of the three numbers is 21:
      $(a - d) + a + (a + d) = 21$
      $3a = 21$
      $a = \frac{21}{3} = 7$
  
  3. From the second given condition, the sum of their squares is 165:
      $(a - d)^2 + a^2 + (a + d)^2 = 165$
  
  4. Expanding the binomials:
      $(a^2 - 2ad + d^2) + a^2 + (a^2 + 2ad + d^2) = 165$
      $3a^2 + 2d^2 = 165$
  
  5. Substituting the value of $a = 7$ into the equation:
      $3(7)^2 + 2d^2 = 165$
      $3(49) + 2d^2 = 165$
      $147 + 2d^2 = 165$
  
  6. Solving for $d$:
      $2d^2 = 165 - 147$
      $2d^2 = 18$
      $d^2 = \frac{18}{2} = 9$
      $d = \pm 3$
  
  7. Finding the numbers for both cases:
      * **Case 1:** When $a = 7$ and $d = 3$
          Numbers: $(7 - 3), 7, (7 + 3) \implies \mathbf{4, 7, 10}$
      * **Case 2:** When $a = 7$ and $d = -3$
          Numbers: $(7 - (-3)), 7, (7 + (-3)) \implies \mathbf{10, 7, 4}$
  
  Result: Therefore, the three numbers are **4, 7, and 10**.
• Q17. Divide 32 into four parts which are in AP such that the ratio of the product of extremes to the product of means is $7 : 15$.
  Answer:
  1. When four parts are in an AP, it is convenient to choose them as $(a - 3d)$, $(a - d)$, $(a + d)$, and $(a + 3d)$, where the common difference is $2d$.
  
  2. From the first condition, the sum of these four parts is 32:
      $(a - 3d) + (a - d) + (a + d) + (a + 3d) = 32$
      $4a = 32$
      $a = \frac{32}{4} = 8$
  
  3. Identify the extremes and the means:
      * Extremes: $(a - 3d)$ and $(a + 3d)$
      * Means: $(a - d)$ and $(a + d)$
  
  4. From the second condition, the ratio of the product of extremes to the product of means is $7 : 15$:
      $\frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{7}{15}$
      $\frac{a^2 - 9d^2}{a^2 - d^2} = \frac{7}{15}$
  
  5. Substitute $a = 8$ into the equation ($a^2 = 64$):
      $\frac{64 - 9d^2}{64 - d^2} = \frac{7}{15}$
  
  6. Cross-multiply to solve for $d$:
      $15(64 - 9d^2) = 7(64 - d^2)$
      $960 - 135d^2 = 448 - 7d^2$
  
  7. Rearranging the terms:
      $960 - 448 = 135d^2 - 7d^2$
      $512 = 128d^2$
      $d^2 = \frac{512}{128} = 4$
      $d = \pm 2$
  
  8. Finding the four parts for $a = 8$ and $d = 2$ (taking $d = 2$):
      * $a - 3d = 8 - 3(2) = 8 - 6 = 2$
      * $a - d = 8 - 2 = 6$
      * $a + d = 8 + 2 = 10$
      * $a + 3d = 8 + 3(2) = 8 + 6 = 14$
  
  Result: Therefore, the four parts are **2, 6, 10, and 14**.
• Q18. Find four numbers in AP whose sum is 20 and the sum of whose squares is 120.
  Answer:
  1. Let the four numbers in AP be $(a - 3d)$, $(a - d)$, $(a + d)$, and $(a + 3d)$, where the common difference is $2d$.
  
  2. From the first condition, the sum of the four numbers is 20:
      $(a - 3d) + (a - d) + (a + d) + (a + 3d) = 20$
      $4a = 20$
      $a = \frac{20}{4} = 5$
  
  3. From the second condition, the sum of their squares is 120:
      $(a - 3d)^2 + (a - d)^2 + (a + d)^2 + (a + 3d)^2 = 120$
  
  4. Expanding the binomials:
      $(a^2 - 6ad + 9d^2) + (a^2 - 2ad + d^2) + (a^2 + 2ad + d^2) + (a^2 + 6ad + 9d^2) = 120$
      $4a^2 + 20d^2 = 120$
  
  5. Substituting the value of $a = 5$ into the equation ($a^2 = 25$):
      $4(25) + 20d^2 = 120$
      $100 + 20d^2 = 120$
  
  6. Solving for $d$:
      $20d^2 = 120 - 100$
      $20d^2 = 20$
      $d^2 = 1 \implies d = \pm 1$
  
  7. Finding the four numbers for $a = 5$ and $d = 1$ (taking $d = 1$):
      * $a - 3d = 5 - 3(1) = 2$
      * $a - d = 5 - 1 = 4$
      * $a + d = 5 + 1 = 6$
      * $a + 3d = 5 + 3(1) = 8$
  
  Result: Therefore, the four numbers are **2, 4, 6, and 8**.
• Q19. A spiral is made up of successive semicircles, with centres alternately at $A$ and $B$, starting with centre at $A$, of radii $0.5\text{ cm}, 1.0\text{ cm}, 1.5\text{ cm}, 2.0\text{ cm}, \dots$ What is the total length of such a spiral made up of thirteen consecutive semicircles? ($\pi = \frac{22}{7}$)
  Answer:
  1. The perimeter (length) of a semicircle with radius $r$ is given by $\pi r$.
  
  2. Let the lengths of the successive semicircles be $L_1, L_2, L_3, L_4, \dots$
      * $L_1 = \pi (0.5) = 0.5\pi$
      * $L_2 = \pi (1.0) = 1.0\pi$
      * $L_3 = \pi (1.5) = 1.5\pi$
      * $L_4 = \pi (2.0) = 2.0\pi$
  
  3. These lengths form an Arithmetic Progression (AP) where:
      * First term ($a$) = $0.5\pi$
      * Common difference ($d$) = $1.0\pi - 0.5\pi = 0.5\pi$
      * Number of terms ($n$) = $13$
  
  4. Using the AP sum formula $S_n = \frac{n}{2}[2a + (n - 1)d]$ to find the total length for 13 semicircles ($S_{13}$):
      $S_{13} = \frac{13}{2}[2(0.5\pi) + (13 - 1)(0.5\pi)]$
      $S_{13} = \frac{13}{2}[\pi + 12(0.5\pi)]$
      $S_{13} = \frac{13}{2}[\pi + 6\pi]$
      $S_{13} = \frac{13}{2} \times 7\pi$
  
  5. Substituting $\pi = \frac{22}{7}$:
      $S_{13} = \frac{13}{2} \times 7 \times \frac{22}{7}$
      $S_{13} = \frac{13}{2} \times 22 = 13 \times 11 = 143\text{ cm}$
  
  Result: Therefore, the total length of the spiral made up of thirteen consecutive semicircles is **$143\text{ cm}$**.
• Q20. The income of a person is ₹3,00,000 in the first year and he receives an increase of ₹10,000 to his income per year for 19 years. Find the total amount he received in 20 years.
  Answer:
  1. The annual incomes of the person in each year form an Arithmetic Progression (AP) where:
      * First term ($a$) = ₹3,00,000 (income in the 1st year)
      * Common difference ($d$) = ₹10,000 (annual increase)
      * Number of terms ($n$) = 20 (total years)
  
  2. We need to find the total amount received in 20 years, which is the sum of the first 20 terms ($S_{20}$).
  
  3. Using the formula for the sum of the first $n$ terms of an AP ($S_n = \frac{n}{2}[2a + (n - 1)d]$):
      $S_{20} = \frac{20}{2}[2(3,00,000) + (20 - 1) \times 10,000]$
  
  4. Simplifying the expression inside the brackets:
      $S_{20} = 10[6,00,000 + 19 \times 10,000]$
      $S_{20} = 10[6,00,000 + 1,90,000]$
      $S_{20} = 10 \times 7,90,000$
      $S_{20} = 79,00,000$
  
  Result: Therefore, the total amount he received in 20 years is **₹79,00,000**.

Academic Repository Disclaimer & எச்சரிக்கை அறிவிப்பு :

English: These model solutions are structured to global academic standards for self-assessment and practice purposes. Students are advised to verify with official textbooks for board examinations.

தமிழ்: இந்த மாதிரி கணிதத் தீர்வுகள் மாணவர்களின் சுய கற்றல் மற்றும் பயிற்சித் திறனுக்காக உலகளாவிய கல்வித் தரத்தில் உருவாக்கப்பட்டுள்ளன. தேர்வுகளுக்குத் தயாராகும் போது உங்கள் அதிகாரப்பூர்வ பாடப்புத்தகத்துடன் சரிபார்க்கவும்.

மேலதிக விவரங்களுக்கு எங்களைத் தொடர்பு கொள்ளவும் → For further details, contact us →

Copyright © 2026 Deepa Maths Academy | Global Examination & Academic Portal. All Rights Reserved.