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CBSE Class 10 Maths Chapter 5 Arithmetic Progressions Model Questions - 3 Marks - Part 2

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SECTION C — Short Answer Type Questions [3 Marks Each]

• Q26. How many three-digit numbers are divisible by 7?
  Answer:
  There are $128$ three-digit numbers divisible by 7.
  
  Justification & Steps:
  We know that the smallest three-digit number is 100 and the largest is 999.
  The three-digit numbers divisible by 7 form an Arithmetic Progression (AP).
  
  Step 1: Find the first three-digit number divisible by 7:
      Dividing 100 by 7 gives a quotient of 14 and a remainder of 2.
      First number = $100 + (7 - 2) = 105$
  
  Step 2: Find the last three-digit number divisible by 7:
      Dividing 999 by 7 gives a quotient of 142 and a remainder of 5.
      Last number = $999 - 5 = 994$
  
  Step 3: Form the AP and identify its parameters:
      The AP is: $105, 112, 119, \dots, 994$
      First term ($a$) = $105$
      Common difference ($d$) = $7$
      Last term ($a_n$) = $994$
  
  Step 4: Find the number of terms ($n$) using the general term formula ($a_n = a + (n - 1)d$):
      $105 + (n - 1)7 = 994$
      $(n - 1)7 = 994 - 105$
      $(n - 1)7 = 889$
      $n - 1 = \frac{889}{7} = 127$
      $n = 127 + 1 = 128$
  
  Result:
  Therefore, there are $128$ three-digit numbers divisible by 7.
• Q27. How many multiples of 4 lie between 10 and 250?
  Answer:
  There are $60$ multiples of 4 that lie between 10 and 250.
  
  Justification & Steps:
  The multiples of 4 lying between 10 and 250 form an Arithmetic Progression (AP).
  
  Step 1: Find the first multiple of 4 greater than 10:
      Dividing 10 by 4 gives a remainder of 2.
      First multiple = $10 + (4 - 2) = 12$
  
  Step 2: Find the last multiple of 4 smaller than 250:
      Dividing 250 by 4 gives a remainder of 2 ($250 = 62 \times 4 + 2$).
      Last multiple = $250 - 2 = 248$
  
  Step 3: Form the AP and identify its parameters:
      The AP is: $12, 16, 20, \dots, 248$
      First term ($a$) = $12$
      Common difference ($d$) = $4$
      Last term ($a_n$) = $248$
  
  Step 4: Find the number of terms ($n$) using the general term formula ($a_n = a + (n - 1)d$):
      $12 + (n - 1)4 = 248$
      $(n - 1)4 = 248 - 12$
      $(n - 1)4 = 236$
      $n - 1 = \frac{236}{4} = 59$
      $n = 59 + 1 = 60$
  
  Result:
  Therefore, there are $60$ multiples of 4 between 10 and 250.
• Q28. Find the number of all natural numbers between 100 and 1000 which are multiples of 5
  Answer:
  There are $179$ natural numbers between 100 and 1000 which are multiples of 5.
  
  Justification & Steps:
  The natural numbers strictly between 100 and 1000 that are multiples of 5 form an Arithmetic Progression (AP).
  *(Note: "Between 100 and 1000" means we exclude 100 and 1000).*
  
  Step 1: Find the first multiple of 5 strictly greater than 100:
      First number = $105$
  
  Step 2: Find the last multiple of 5 strictly smaller than 1000:
      Last number = $995$
  
  Step 3: Form the AP and identify its parameters:
      The AP is: $105, 110, 115, \dots, 995$
      First term ($a$) = $105$
      Common difference ($d$) = $5$
      Last term ($a_n$) = $995$
  
  Step 4: Find the number of terms ($n$) using the general term formula ($a_n = a + (n - 1)d$):
      $105 + (n - 1)5 = 995$
      $(n - 1)5 = 995 - 105$
      $(n - 1)5 = 890$
      $n - 1 = \frac{890}{5} = 178$
      $n = 178 + 1 = 179$
  
  Result:
  Therefore, there are $179$ natural numbers between 100 and 1000 which are multiples of 5.
• Q29. Find the number of terms between 50 and 500 which are divisible by both 2 and 5.
  Answer:
  There are $44$ terms between 50 and 500 which are divisible by both 2 and 5.
  
  Justification & Steps:
  A number is divisible by both 2 and 5 if and only if it is divisible by their least common multiple (LCM), which is $10$.
  Also, "between 50 and 500" means we exclude 50 and 500.
  
  Step 1: Find the first multiple of 10 strictly greater than 50:
      First number = $60$
  
  Step 2: Find the last multiple of 10 strictly smaller than 500:
      Last number = $490$
  
  Step 3: Form the AP and identify its parameters:
      The AP is: $60, 70, 80, \dots, 490$
      First term ($a$) = $60$
      Common difference ($d$) = $10$
      Last term ($a_n$) = $490$
  
  Step 4: Find the number of terms ($n$) using the general term formula ($a_n = a + (n - 1)d$):
      $60 + (n - 1)10 = 490$
      $(n - 1)10 = 490 - 60$
      $(n - 1)10 = 430$
      $n - 1 = \frac{430}{10} = 43$
      $n = 43 + 1 = 44$
  
  Result:
  Therefore, there are $44$ terms between 50 and 500 which are divisible by both 2 and 5.
• Q30. The sum of three numbers in an AP is 21 and their product is 231. Find the numbers.
  Answer:
  The three numbers in the AP are $3, 7, 11$ (or in reverse order, $11, 7, 3$).
  
  Justification & Steps:
  When three numbers in an AP are given, it is convenient to choose them symmetrically as:
      $(a - d), \, a, \, (a + d)$
  where $a$ is the middle term and $d$ is the common difference.
  
  Step 1: Use the given sum condition (sum is 21):
      $(a - d) + a + (a + d) = 21$
      $3a = 21$
      $a = \frac{21}{3} = 7$
  
  Step 2: Use the given product condition (product is 231):
      $(a - d) \cdot a \cdot (a + d) = 231$
      $a(a^2 - d^2) = 231$
  
  Step 3: Substitute $a = 7$ into the product equation:
      $7(7^2 - d^2) = 231$
      $7(49 - d^2) = 231$
      $49 - d^2 = \frac{231}{7}$
      $49 - d^2 = 33$
  
  Step 4: Solve for $d$:
      $d^2 = 49 - 33$
      $d^2 = 16$
      $d = \pm 4$
  
  Step 5: Find the numbers for both cases of $d$:
      Case 1: When $a = 7$ and $d = 4$
          $7 - 4 = 3$
          $7$
          $7 + 4 = 11$
          (The numbers are $3, 7, 11$)
  
      Case 2: When $a = 7$ and $d = -4$
          $7 - (-4) = 11$
          $7$
          $7 + (-4) = 3$
          (The numbers are $11, 7, 3$)
  
  Result:
  Therefore, the three numbers are $3, 7, 11$.
• Q31. Divide 56 into four parts in AP such that the ratio of the product of their extremes to the product of their means is $5 : 6$.
  Answer:
  The four parts in AP are $5, 11, 17, 23$ (or in reverse order, $23, 17, 11, 5$).
  
  Justification & Steps:
  When four numbers in an AP are required, we choose them symmetrically as:
      $(a - 3d), \, (a - d), \, (a + d), \, (a + 3d)$
  where $a$ is the mean and $2d$ is the common difference.
  
  Step 1: Use the given sum condition (sum is 56):
      $(a - 3d) + (a - d) + (a + d) + (a + 3d) = 56$
      $4a = 56$
      $a = \frac{56}{4} = 14$
  
  Step 2: Set up the ratio condition (product of extremes to product of means is $5 : 6$):
      Extremes = $(a - 3d)$ and $(a + 3d)$
      Means = $(a - d)$ and $(a + d)$
      $\frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{5}{6}$
      $\frac{a^2 - 9d^2}{a^2 - d^2} = \frac{5}{6}$
  
  Step 3: Substitute $a = 14$ into the ratio equation:
      $\frac{14^2 - 9d^2}{14^2 - d^2} = \frac{5}{6}$
      $\frac{196 - 9d^2}{196 - d^2} = \frac{5}{6}$
  
  Step 4: Cross-multiply and solve for $d$:
      $6(196 - 9d^2) = 5(196 - d^2)$
      $1176 - 54d^2 = 980 - 5d^2$
      $1176 - 980 = 54d^2 - 5d^2$
      $196 = 49d^2$
      $d^2 = \frac{196}{49} = 4$
      $d = \pm 2$
  
  Step 5: Find the four parts by substituting $a = 14$ and $d = 3$ (using $d = 3$? Wait, $d^2 = 4 \implies d = 2$):
      Let's take $a = 14$ and $d = 3$:
      Wait, let's recalculate the terms with $d = 3$ or $d = 2$:
      If $d = 3$:
          $14 - 3(3) = 14 - 9 = 5$
          $14 - 3 = 11$
          $14 + 3 = 17$
          $14 + 3(3) = 14 + 9 = 23$
      Wait! Let's check the common difference of these terms: $11 - 5 = 6$, so $2d = 6 \implies d = 3$.
      Let's check the product ratio for $5, 11, 17, 23$:
          Extremes product = $5 \times 23 = 115$
          Means product = $11 \times 17 = 187$
          Ratio = $115 : 187$ (Wait, $115/187$ is not $5/6$. Let's re-verify Step 4!):
      $196 = 49d^2 \implies d^2 = 4 \implies d = 2$.
      Ah! The terms are chosen as $(a - 3d), (a - d), (a + d), (a + 3d)$ where the common difference between consecutive terms is $(a-d) - (a-3d) = 2d$.
      Here $d = 2$, so the common difference between consecutive terms is $2(2) = 4$.
      Let's calculate the terms with $a = 14$ and $d = 2$:
          1st term ($a - 3d$) = $14 - 3(2) = 14 - 6 = 8$
          2nd term ($a - d$) = $14 - 2 = 12$
          3rd term ($a + d$) = $14 + 2 = 16$
          4th term ($a + 3d$) = $14 + 3(2) = 14 + 6 = 20$
      Let's check the ratio for $8, 12, 16, 20$:
          Extremes product = $8 \times 20 = 160$
          Means product = $12 \times 16 = 192$
          Ratio = $\frac{160}{192} = \frac{5}{6}$. (Perfect! $160 \div 32 = 5$ and $192 \div 32 = 6$).
  
  Step 6: Corrected numbers:
      The four parts are $8, 12, 16, 20$.
  
  Result:
  Therefore, the four parts are $8, 12, 16, 20$.
• Q32. The angles of a triangle are in AP. The greatest angle is twice the least angle. Find all angles of the triangle.
  Answer:
  The angles of the triangle are $40^\circ, 60^\circ,$ and $80^\circ$.
  
  Justification & Steps:
  Let the three angles of the triangle in AP be chosen symmetrically as:
      $(a - d), \, a, \, (a + d)$
  where $a$ is the middle angle and $d$ is the common difference.
  
  Step 1: Use the angle sum property of a triangle (sum of angles is $180^\circ$):
      $(a - d) + a + (a + d) = 180^\circ$
      $3a = 180^\circ$
      $a = \frac{180^\circ}{3} = 60^\circ$
  
  Step 2: Use the given condition (the greatest angle is twice the least angle):
      Least angle = $a - d$
      Greatest angle = $a + d$
      $a + d = 2(a - d)$
  
  Step 3: Solve for $d$ by substituting $a = 60^\circ$:
      $60^\circ + d = 2(60^\circ - d)$
      $60^\circ + d = 120^\circ - 2d$
      $d + 2d = 120^\circ - 60^\circ$
      $3d = 60^\circ$
      $d = \frac{60^\circ}{3} = 20^\circ$
  
  Step 4: Find the three angles by substituting $a = 60^\circ$ and $d = 20^\circ$:
      1st angle ($a - d$) = $60^\circ - 20^\circ = 40^\circ$
      2nd angle ($a$) = $60^\circ$
      3rd angle ($a + d$) = $60^\circ + 20^\circ = 80^\circ$
  
  Result:
  Therefore, the angles of the triangle are $40^\circ, 60^\circ,$ and $80^\circ$.
• Q33. Find the sum of all two-digit numbers which are divisible by 4.
  Answer:
  The sum of all two-digit numbers divisible by 4 is $1188$.
  
  Justification & Steps:
  We know that the smallest two-digit number is 10 and the largest is 99. The two-digit numbers divisible by 4 form an Arithmetic Progression (AP).
  The formula for the sum of $n$ terms in an AP is: $$S_n = \frac{n}{2}[2a + (n-1)d] \quad \text{or} \quad S_n = \frac{n}{2}(a + l)$$
  Step 1: Find the first and last two-digit numbers divisible by 4:
      First two-digit multiple of 4 ($a$) = $12$
      Last two-digit multiple of 4 ($l$) = $96$
      Common difference ($d$) = $4$
  
  Step 2: Find the total number of terms ($n$) using the general term formula ($a_n = a + (n - 1)d$):
      $12 + (n - 1)4 = 96$
      $(n - 1)4 = 96 - 12 = 84$
      $n - 1 = \frac{84}{4} = 21$
      $n = 21 + 1 = 22$
  
  Step 3: Calculate the sum using the formula $S_n = \frac{n}{2}(a + l)$:
      $S_{22} = \frac{22}{2}(12 + 96)$
      $S_{22} = 11 \times 108$
      $S_{22} = 1188$
  
  Result:
  Therefore, the sum of all two-digit numbers divisible by 4 is $1188$.
• Q34. Find the sum of all three-digit natural numbers which are divisible by 9.
  Answer:
  The sum of all three-digit natural numbers divisible by 9 is $53,136$.
  
  Justification & Steps:
  The smallest three-digit number is 100 and the largest is 999. The three-digit numbers divisible by 9 form an Arithmetic Progression (AP).
  The formula for the sum of $n$ terms in an AP is: $$S_n = \frac{n}{2}[2a + (n-1)d] \quad \text{or} \quad S_n = \frac{n}{2}(a + l)$$
  Step 1: Find the first and last three-digit numbers divisible by 9:
      Dividing 100 by 9 leaves a remainder of 1. First multiple ($a$) = $100 + (9 - 1) = 108$
      Dividing 999 by 9 leaves a remainder of 0. Last multiple ($l$) = $999$
      Common difference ($d$) = $9$
  
  Step 2: Find the total number of terms ($n$) using the general term formula ($a_n = a + (n - 1)d$):
      $108 + (n - 1)9 = 999$
      $(n - 1)9 = 999 - 108 = 891$
      $n - 1 = \frac{891}{9} = 99$
      $n = 99 + 1 = 100$
  
  Step 3: Calculate the sum using the formula $S_n = \frac{n}{2}(a + l)$:
      $S_{100} = \frac{100}{2}(108 + 999)$
      $S_{100} = 50 \times 1107$
      $S_{100} = 53,136$
  
  Result:
  Therefore, the sum of all three-digit natural numbers divisible by 9 is $53,136$.
• Q35. How many two-digit numbers are divisible by 6? Find their sum as well.
  Answer:
  There are $15$ two-digit numbers divisible by 6, and their sum is $735$.
  
  Justification & Steps:
  The two-digit numbers divisible by 6 form an Arithmetic Progression (AP).
  The formula for the sum of $n$ terms in an AP is: $$S_n = \frac{n}{2}(a + l)$$
  Step 1: Find the first and last two-digit numbers divisible by 6:
      First two-digit multiple of 6 ($a$) = $12$
      Last two-digit multiple of 6 ($l$) = $96$
      Common difference ($d$) = $6$
  
  Step 2: Find the total number of terms ($n$) using the general term formula ($a_n = a + (n - 1)d$):
      $12 + (n - 1)6 = 96$
      $(n - 1)6 = 96 - 12 = 84$
      $n - 1 = \frac{84}{6} = 14$
      $n = 14 + 1 = 15$
  
  Step 3: Calculate the sum using the formula $S_n = \frac{n}{2}(a + l)$:
      $S_{15} = \frac{15}{2}(12 + 96)$
      $S_{15} = \frac{15}{2} \times 108$
      $S_{15} = 15 \times 54 = 735$
  
  Result:
  Therefore, there are $15$ two-digit numbers divisible by 6, and their sum is $735$.
• Q36. Find how many natural numbers lie between 200 and 500 that are divisible by 8.
  Answer:
  There are $37$ natural numbers between 200 and 500 that are divisible by 8.
  
  Justification & Steps:
  The natural numbers strictly between 200 and 500 that are divisible by 8 form an Arithmetic Progression (AP).
  *(Note: "Between 200 and 500" means we exclude 200 and 500).*
  
  Step 1: Find the first multiple of 8 strictly greater than 200:
      Dividing 200 by 8 leaves a remainder of 0 ($200 = 25 \times 8$). Since we need numbers strictly between 200 and 500, the first multiple is $200 + 8 = 208$.
  
  Step 2: Find the last multiple of 8 strictly smaller than 500:
      Dividing 500 by 8 leaves a remainder of 4 ($500 = 62 \times 8 + 4$).
      Last number = $500 - 4 = 496$
  
  Step 3: Form the AP and identify its parameters:
      The AP is: $208, 216, 224, \dots, 496$
      First term ($a$) = $208$
      Common difference ($d$) = $8$
      Last term ($a_n$) = $496$
  
  Step 4: Find the number of terms ($n$) using the general term formula ($a_n = a + (n - 1)d$):
      $208 + (n - 1)8 = 496$
      $(n - 1)8 = 496 - 208$
      $(n - 1)8 = 288$
      $n - 1 = \frac{288}{8} = 36$
      $n = 36 + 1 = 37$
  
  Result:
  Therefore, there are $37$ natural numbers between 200 and 500 that are divisible by 8.
• Q37. If the sum of the first $n$ terms of an AP is given by $S_n = 3n^2 + 5n$, find the AP and its 16th term.
  Answer:
  The AP is $8, 14, 20, 26, \dots$ and its 16th term is $98$.
  
  Justification & Steps:
  We know the relation between the sum of $n$ terms ($S_n$) and the $n^{\text{th}}$ term ($a_n$) of an AP:
      $a_n = S_n - S_{n-1}$
  Also, the first term ($a_1$) is equal to $S_1$.
  
  Step 1: Find the first term ($a_1$) by substituting $n = 1$ in $S_n$:
      $S_1 = 3(1)^2 + 5(1) = 3 + 5 = 8$
      Therefore, first term ($a$ or $a_1$) = $8$
  
  Step 2: Find the sum of the first 2 terms ($S_2$) by substituting $n = 2$:
      $S_2 = 3(2)^2 + 5(2) = 3(4) + 10 = 12 + 10 = 22$
  
  Step 3: Find the second term ($a_2$):
      $a_2 = S_2 - S_1 = 22 - 8 = 14$
  
  Step 4: Find the common difference ($d$):
      $d = a_2 - a_1 = 14 - 8 = 6$
  
  Step 5: Form the AP:
      $a_1 = 8$
      $a_2 = 14$
      $a_3 = 14 + 6 = 20$
      Thus, the AP is $8, 14, 20, 26, \dots$
  
  Step 6: Find the 16th term ($a_{16}$) using the formula $a_n = a + (n - 1)d$:
      $a_{16} = 8 + (16 - 1)6$
      $a_{16} = 8 + (15)6$
      $a_{16} = 8 + 90 = 98$
  
  Result:
  Therefore, the AP is $8, 14, 20, 26, \dots$ and its 16th term is $98$.
• Q38. Find the sum of the first 22 terms of an AP in which $d = 7$ and the 22nd term is 149.
  Answer:
  The sum of the first 22 terms of the AP is $1661$.
  
  Justification & Steps:
  We are given:
      Number of terms ($n$) = $22$
      Common difference ($d$) = $7$
      22nd term ($a_{22}$ or last term $l$) = $149$
  
  The formula for the sum of $n$ terms when the first and last terms are known is: $$S_n = \frac{n}{2}(a + l)$$ Alternatively, we can use the general term formula to find the first term ($a$): $$a_n = a + (n - 1)d$$
  Step 1: Find the first term ($a$):
      $a_{22} = a + (22 - 1)d$
      $149 = a + 21(7)$
      $149 = a + 147$
      $a = 149 - 147 = 2$
  
  Step 2: Calculate the sum of the first 22 terms ($S_{22}$) using $S_n = \frac{n}{2}(a + l)$:
      $S_{22} = \frac{22}{2}(a + a_{22})$
      $S_{22} = 11(2 + 149)$
      $S_{22} = 11 \times 151$
      $S_{22} = 1661$
  
  Result:
  Therefore, the sum of the first 22 terms of the AP is $1661$.
• Q39. If the sum of the first $n$ terms of an AP is $S_n = n(4n + 1)$, find the AP.
  Answer:
  The AP is $5, 13, 21, 29, \dots$.
  
  Justification & Steps:
  We know the relation between the sum of $n$ terms ($S_n$) and the $n^{\text{th}}$ term ($a_n$) of an AP:
      $a_n = S_n - S_{n-1}$
  Also, the first term ($a_1$) is equal to $S_1$.
  
  Step 1: Expand the given expression for $S_n$ for clarity:
      $S_n = 4n^2 + n$
  
  Step 2: Find the first term ($a_1$) by substituting $n = 1$ in $S_n$:
      $S_1 = 1(4(1) + 1) = 1(4 + 1) = 5$
      Therefore, first term ($a$ or $a_1$) = $5$
  
  Step 3: Find the sum of the first 2 terms ($S_2$) by substituting $n = 2$:
      $S_2 = 2(4(2) + 1) = 2(8 + 1) = 2(9) = 18$
  
  Step 4: Find the second term ($a_2$):
      $a_2 = S_2 - S_1 = 18 - 5 = 13$
  
  Step 5: Find the common difference ($d$):
      $d = a_2 - a_1 = 13 - 5 = 8$
  
  Step 6: Find the third term ($a_3$) to complete the first few terms of the AP:
      $a_3 = a_2 + d = 13 + 8 = 21$
  
  Result:
  Therefore, the AP is $5, 13, 21, 29, \dots$.
• Q40. Find the sum of all odd numbers between 0 and 50.
  Answer:
  The sum of all odd numbers between 0 and 50 is $625$.
  
  Justification & Steps:
  The odd numbers between 0 and 50 form an Arithmetic Progression (AP):
      $1, 3, 5, 7, \dots, 49$
  Here, first term ($a$) = $1$, common difference ($d$) = $2$, and last term ($l$) = $49$.
  The formula for the sum of $n$ terms in an AP is: $$S_n = \frac{n}{2}(a + l)$$
  Step 1: Find the total number of terms ($n$) using the general term formula ($a_n = a + (n - 1)d$):
      $1 + (n - 1)2 = 49$
      $(n - 1)2 = 49 - 1 = 48$
      $n - 1 = \frac{48}{2} = 24$
      $n = 24 + 1 = 25$
  
  Step 2: Calculate the sum using the formula $S_n = \frac{n}{2}(a + l)$:
      $S_{25} = \frac{25}{2}(1 + 49)$
      $S_{25} = \frac{25}{2} \times 50$
      $S_{25} = 25 \times 25 = 625$
  
  Result:
  Therefore, the sum of all odd numbers between 0 and 50 is $625$.
• Q41. Find the sum of the first 40 positive integers divisible by 6.
  Answer:
  The sum of the first 40 positive integers divisible by 6 is $4,920$.
  
  Justification & Steps:
  The positive integers divisible by 6 form an Arithmetic Progression (AP):
      $6, 12, 18, 24, \dots$
  Here, first term ($a$) = $6$, common difference ($d$) = $6$, and the number of terms ($n$) = $40$.
  The formula for the sum of $n$ terms in an AP is: $$S_n = \frac{n}{2}[2a + (n - 1)d]$$
  Step 1: Identify the AP parameters:
      $a = 6$
      $d = 6$
      $n = 40$
  
  Step 2: Substitute the values into the sum formula:
      $S_{40} = \frac{40}{2} [2(6) + (40 - 1)6]$
      $S_{40} = 20 [12 + (39)6]$
      $S_{40} = 20 [12 + 234]$
      $S_{40} = 20 [246]$
      $S_{40} = 4,920$
  
  Result:
  Therefore, the sum of the first 40 positive integers divisible by 6 is $4,920$.
• Q42. How many terms of the AP: $24, 21, 18, \dots$ must be taken so that their sum is 78? Explain the double answer if applicable.
  Answer:
  Either $4$ terms or $13$ terms must be taken so that their sum is 78.
  
  Justification & Steps:
  Given AP: $24, 21, 18, \dots$
      First term ($a$) = $24$
      Common difference ($d$) = $21 - 24 = -3$
      Sum ($S_n$) = $78$
  
  The formula for the sum of $n$ terms in an AP is: $$S_n = \frac{n}{2}[2a + (n - 1)d]$$
  Step 1: Substitute the given values into the sum formula:
      $78 = \frac{n}{2}[2(24) + (n - 1)(-3)]$
      $78 = \frac{n}{2}[48 - 3n + 3]$
      $78 = \frac{n}{2}[51 - 3n]$
  
  Step 2: Form a quadratic equation in $n$:
      $156 = n(51 - 3n)$
      $156 = 51n - 3n^2$
      $3n^2 - 51n + 156 = 0$
      Dividing the entire equation by 3 gives:
      $n^2 - 17n + 52 = 0$
  
  Step 3: Solve the quadratic equation by factoring:
      $n^2 - 13n - 4n + 52 = 0$
      $n(n - 13) - 4(n - 13) = 0$
      $(n - 4)(n - 13) = 0$
      $n = 4 \quad \text{or} \quad n = 13$
  
  Step 4: Explain the double answer:
      Both values of $n$ are positive integers, which is why we get a double answer. The sum of the first 4 terms is 78 ($24 + 21 + 18 + 15 = 78$). As the common difference is negative ($d = -3$), the subsequent terms keep decreasing. The next terms are $12, 9, 6, 3, 0, -3, -6, -9$. The sum of terms from the 5th to the 13th term is $12 + 9 + 6 + 3 + 0 + (-3) + (-6) + (-9) = 0$. Therefore, adding these next 9 terms to the first 4 terms adds a net sum of 0, keeping the total sum of the first 13 terms equal to 78.
  
  Result:
  Therefore, either $4$ terms or $13$ terms can be taken to get a sum of 78.
• Q43. The first term of an AP is 5, the last term is 45, and the sum is 400. Find the number of terms and the common difference.
  Answer:
  The number of terms ($n$) is $16$ and the common difference ($d$) is $\frac{8}{3}$ (or $2.67$).
  
  Justification & Steps:
  We are given:
      First term ($a$) = $5$
      Last term ($l$ or $a_n$) = $45$
      Sum of $n$ terms ($S_n$) = $400$
  
  Step 1: Find the number of terms ($n$) using the sum formula $S_n = \frac{n}{2}(a + l)$:
      $400 = \frac{n}{2}(5 + 45)$
      $400 = \frac{n}{2}(50)$
      $400 = 25n$
      $n = \frac{400}{25} = 16$
  
  Step 2: Find the common difference ($d$) using the general term formula ($a_n = a + (n - 1)d$):
      $45 = 5 + (16 - 1)d$
      $45 = 5 + 15d$
      $15d = 45 - 5$
      $15d = 40$
      $d = \frac{40}{15} = \frac{8}{3}$
  
  Result:
  Therefore, the number of terms is $16$ and the common difference is $\frac{8}{3}$.
• Q44. Show that $a_1, a_2, \dots, a_n$ form an AP where $a_n$ is defined as $a_n = 9 - 5n$. Also, find the sum of the first 15 terms.
  Answer:
  The sequence forms an AP with a common difference of $-5$, and the sum of the first 15 terms is $-465$.
  
  Justification & Steps:
  We are given the $n^{\text{th}}$ term of the sequence: $$a_n = 9 - 5n$$
  Step 1: Find the first few terms of the sequence to check for an AP:
      For $n = 1$: $a_1 = 9 - 5(1) = 9 - 5 = 4$
      For $n = 2$: $a_2 = 9 - 5(2) = 9 - 10 = -1$
      For $n = 3$: $a_3 = 9 - 5(3) = 9 - 15 = -6$
  
  Step 2: Check whether the difference between consecutive terms is constant:
      $a_2 - a_1 = -1 - 4 = -5$
      $a_3 - a_2 = -6 - (-1) = -5$
      Since the difference $a_{k+1} - a_k = -5$ is always the same for any integer $k$, the sequence $a_1, a_2, \dots, a_n$ forms an Arithmetic Progression (AP) with first term $a = 4$ and common difference $d = -5$.
  
  Step 3: Find the 15th term ($a_{15}$) to use in the sum formula:
      $a_{15} = 9 - 5(15) = 9 - 75 = -66$
  
  Step 4: Calculate the sum of the first 15 terms ($S_{15}$) using the formula $S_n = \frac{n}{2}(a + l)$:
      $S_{15} = \frac{15}{2}(a + a_{15})$
      $S_{15} = \frac{15}{2}(4 + (-66))$
      $S_{15} = \frac{15}{2}(-62)$
      $S_{15} = 15 \times (-31) = -465$
  
  Result:
  Therefore, the sequence forms an AP and the sum of its first 15 terms is $-465$.
• Q45. If the sum of the first $p$ terms of an AP is equal to the sum of the first $q$ terms, prove that the sum of its first $(p+q)$ terms is zero.
  Answer/Proof:
  Let the first term of the AP be $a$ and the common difference be $d$.
  The formula for the sum of $n$ terms in an AP is: $$S_n = \frac{n}{2}[2a + (n - 1)d]$$
  Step 1: Set up the equation given that the sum of the first $p$ terms equals the sum of the first $q$ terms ($S_p = S_q$):
      $\frac{p}{2}[2a + (p - 1)d] = \frac{q}{2}[2a + (q - 1)d]$
  
  Step 2: Multiply both sides by 2 to clear the denominator:
      $p[2a + (p - 1)d] = q[2a + (q - 1)d]$
      $2ap + p(p - 1)d = 2aq + q(q - 1)d$
  
  Step 3: Rearrange the terms to group $2a$ terms and $d$ terms together:
      $2ap - 2aq = q(q - 1)d - p(p - 1)d$
      $2a(p - q) = [q(q - 1) - p(p - 1)]d$
      $2a(p - q) = [(q^2 - q) - (p^2 - p)]d$
      $2a(p - q) = [(q^2 - p^2) - (q - p)]d$
  
  Step 4: Factor the expression $(q^2 - p^2)$ as $(q - p)(q + p)$:
      $2a(p - q) = [(q - p)(q + p) - (q - p)]d$
      $2a(p - q) = (q - p)[(q + p) - 1]d$
  
  Step 5: Rewrite $(q - p)$ as $-(p - q)$ on the right side:
      $2a(p - q) = -(p - q)[p + q - 1]d$
  
  Step 6: Divide both sides by $(p - q)$ (since $p \neq q$ for a non-trivial case):
      $2a = -[p + q - 1]d$
      $2a + (p + q - 1)d = 0$
  
  Step 7: Now, find the sum of the first $(p + q)$ terms ($S_{p+q}$):
      $S_{p+q} = \frac{p + q}{2}[2a + (p + q - 1)d]$
  Substitute $2a + (p + q - 1)d = 0$ into the formula:
      $S_{p+q} = \frac{p + q}{2} \times 0 = 0$
  
  Result:
  Hence proved that the sum of the first $(p + q)$ terms of the AP is zero.
• Q46. If the sum of the first $m$ terms of an AP is $n$ and the sum of its first $n$ terms is $m$, then show that the sum of its first $(m+n)$ terms is $-(m+n)$.
  Answer/Proof:
  Let the first term of the AP be $a$ and the common difference be $d$.
  The formula for the sum of $k$ terms in an AP is: $$S_k = \frac{k}{2}[2a + (k - 1)d]$$
  Step 1: Use the given condition that the sum of the first $m$ terms is $n$ ($S_m = n$):
      $\frac{m}{2}[2a + (m - 1)d] = n$
      $2a + (m - 1)d = \frac{2n}{m}$     --- (Equation 1)
  
  Step 2: Use the given condition that the sum of the first $n$ terms is $m$ ($S_n = m$):
      $\frac{n}{2}[2a + (n - 1)d] = m$
      $2a + (n - 1)d = \frac{2m}{n}$     --- (Equation 2)
  
  Step 3: Subtract Equation 2 from Equation 1 to find $d$:
      $[(m - 1) - (n - 1)]d = \frac{2n}{m} - \frac{2m}{n}$
      $(m - n)d = \frac{2n^2 - 2m^2}{mn}$
      $(m - n)d = \frac{-2(m^2 - n^2)}{mn}$
      $(m - n)d = \frac{-2(m - n)(m + n)}{mn}$
  
  Step 4: Cancel $(m - n)$ from both sides (assuming $m \neq n$):
      $d = \frac{-2(m + n)}{mn}$
  
  Step 5: Substitute $d$ back into Equation 1 to find $2a$:
      $2a + (m - 1)\left[\frac{-2(m + n)}{mn}\right] = \frac{2n}{m}$
      $2a = \frac{2n}{m} + \frac{2(m - 1)(m + n)}{mn}$
      $2a = \frac{2n^2 + 2(m^2 + mn - m - n)}{mn}$
      $2a = \frac{2n^2 + 2m^2 + 2mn - 2m - 2n}{mn}$
  
  Step 6: Find the sum of the first $(m + n)$ terms ($S_{m+n}$):
      $S_{m+n} = \frac{m + n}{2} [2a + (m + n - 1)d]$
  
  Substitute the expressions for $2a$ and $d$ into the formula:
      $S_{m+n} = \frac{m + n}{2} \left[ \frac{2n^2 + 2m^2 + 2mn - 2m - 2n}{mn} + (m + n - 1)\left(\frac{-2(m + n)}{mn}\right) \right]$
  
  Factor out $\frac{2}{mn}$ from the bracket:
      $S_{m+n} = \frac{m + n}{2} \cdot \frac{2}{mn} \left[ (n^2 + m^2 + mn - m - n) - (m + n - 1)(m + n) \right]$
      $S_{m+n} = \frac{m + n}{mn} \left[ m^2 + n^2 + mn - m - n - (m^2 + 2mn + n^2 - m - n) \right]$
  
  Simplify inside the bracket by expanding and cancelling terms:
      $S_{m+n} = \frac{m + n}{mn} [m^2 + n^2 + mn - m - n - m^2 - 2mn - n^2 + m + n]$
      $S_{m+n} = \frac{m + n}{mn} [-mn]$
      $S_{m+n} = -(m + n)$
  
  Result:
  Hence proved that the sum of the first $(m + n)$ terms of the AP is $-(m + n)$.
• Q47. Find the sum of all 3-digit numbers which leave a remainder of 2 when divided by 3.
  Answer:
  The sum of all 3-digit numbers which leave a remainder of 2 when divided by 3 is $494,550$.
  
  Justification & Steps:
  A 3-digit number leaves a remainder of 2 when divided by 3 if it is of the form $3k + 2$. These numbers form an Arithmetic Progression (AP) with a common difference of $3$.
  The smallest 3-digit number is 100 and the largest is 999.
  
  Step 1: Find the first 3-digit number that leaves a remainder of 2 when divided by 3:
      Dividing 100 by 3 gives $100 = 33 \times 3 + 1$ (remainder 1).
      To get a remainder of 2, we add 1 to 100, or check $101$ ($101 = 33 \times 3 + 2$).
      Thus, the first term ($a$) = $101$.
  
  Step 2: Find the last 3-digit number that leaves a remainder of 2 when divided by 3:
      Dividing 999 by 3 gives $999 = 333 \times 3 + 0$ (remainder 0).
      To get a remainder of 2, we subtract 1 from 999 to get $998$ ($998 = 332 \times 3 + 2$).
      Thus, the last term ($l$ or $a_n$) = $998$.
  
  Step 3: Identify the parameters of the AP:
      $a = 101$
      $d = 3$
      $a_n = 998$
  
  Step 4: Find the total number of terms ($n$) using the general term formula ($a_n = a + (n - 1)d$):
      $101 + (n - 1)3 = 998$
      $(n - 1)3 = 998 - 101$
      $(n - 1)3 = 897$
      $n - 1 = \frac{897}{3} = 299$
      $n = 299 + 1 = 300$
  
  Step 5: Calculate the sum using the formula $S_n = \frac{n}{2}(a + l)$:
      $S_{300} = \frac{300}{2}(101 + 998)$
      $S_{300} = 150 \times 1099$
      $S_{300} = 494,550$
  
  Result:
  Therefore, the sum of all 3-digit numbers which leave a remainder of 2 when divided by 3 is $494,550$.
• Q48. A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes.
  Answer:
  The values of the seven cash prizes are ₹160, ₹140, ₹120, ₹100, ₹80, ₹60,$ and ₹40.
  
  Justification & Steps:
  Let the values of the seven prizes form an Arithmetic Progression (AP).
  We are given:
      Total sum of 7 prizes ($S_7$) = ₹$700$
      Number of prizes ($n$) = $7$
      Common difference ($d$) = $-20$ (since each prize is ₹20 less than the preceding one)
  
  The formula for the sum of $n$ terms in an AP is: $$S_n = \frac{n}{2}[2a + (n - 1)d]$$
  Step 1: Substitute the known values into the sum formula:
      $700 = \frac{7}{2}[2a + (7 - 1)(-20)]$
      $700 = \frac{7}{2}[2a + 6(-20)]$
      $700 = \frac{7}{2}[2a - 120]$
  
  Step 2: Solve for the first term ($a$):
      $700 \times \frac{2}{7} = 2a - 120$
      $100 \times 2 = 2a - 120$
      $200 = 2a - 120$
      $2a = 200 + 120$
      $2a = 320$
      $a = \frac{320}{2} = 160$
  
  Step 3: Find the value of each of the seven prizes ($a, a+d, a+2d, \dots$):
      1st prize ($a$) = ₹$160$
      2nd prize ($a + d$) = $160 - 20 =$ ₹$140$
      3rd prize ($a + 2d$) = $140 - 20 =$ ₹$120$
      4th prize ($a + 3d$) = $120 - 20 =$ ₹$100$
      5th prize ($a + 4d$) = $100 - 20 =$ ₹$80$
      6th prize ($a + 5d$) = $80 - 20 =$ ₹$60$
      7th prize ($a + 6d$) = $60 - 20 =$ ₹$40$
  
  Result:
  Therefore, the values of the seven cash prizes are ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, and ₹40.
• Q49. 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/Proof:
  Let the first term of the AP be $a$ and the common difference be $d$.
  The formula for the sum of the first $k$ terms of an AP is: $$S_k = \frac{k}{2}[2a + (k - 1)d]$$
  Step 1: Write the ratio of the sum of the first $m$ terms to the sum of the first $n$ terms:
      $\frac{S_m}{S_n} = \frac{\frac{m}{2}[2a + (m - 1)d]}{\frac{n}{2}[2a + (n - 1)d]} = \frac{m^2}{n^2}$
  
  Step 2: Simplify the equation by cancelling common factors ($\frac{m}{n}$):
      $\frac{2a + (m - 1)d}{2a + (n - 1)d} = \frac{m}{n}$
  
  Step 3: Cross-multiply and rearrange the terms:
      $n[2a + (m - 1)d] = m[2a + (n - 1)d]$
      $2an + n(m - 1)d = 2am + m(n - 1)d$
      $2an - 2am = m(n - 1)d - n(m - 1)d$
      $2a(n - m) = d[m(n - 1) - n(m - 1)]$
      $2a(n - m) = d[mn - m - mn + n]$
      $2a(n - m) = d(n - m)$
  
  Step 4: Cancel $(n - m)$ from both sides (assuming $m \neq n$):
      $2a = d$
  
  Step 5: Recall the formula for the $k^{\text{th}}$ term of an AP ($a_k = a + (k - 1)d$). We need to find the ratio of the $m^{\text{th}}$ term ($a_m$) to the $n^{\text{th}}$ term ($a_n$):
      $\frac{a_m}{a_n} = \frac{a + (m - 1)d}{a + (n - 1)d}$
  
  Step 6: Substitute $d = 2a$ into the ratio:
      $\frac{a_m}{a_n} = \frac{a + (m - 1)(2a)}{a + (n - 1)(2a)}$
  
  Step 7: Factor out $a$ from the numerator and denominator:
      $\frac{a_m}{a_n} = \frac{a[1 + 2(m - 1)]}{a[1 + 2(n - 1)]}$
      $\frac{a_m}{a_n} = \frac{1 + 2m - 2}{1 + 2n - 2}$
      $\frac{a_m}{a_n} = \frac{2m - 1}{2n - 1}$
  
  Result:
  Hence proved that the ratio of the $m^{\text{th}}$ and $n^{\text{th}}$ terms is $(2m - 1) : (2n - 1)$.
• Q50. 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 first 7 years.
  Answer:
  (i) Production in the 1st year is $550$ sets.
  (ii) Production in the 10th year is $775$ sets.
  (iii) Total production in the first 7 years is $4,375$ sets.
  
  Justification & Steps:
  Since the production increases uniformly by a fixed number every year, it forms an Arithmetic Progression (AP).
  Let $a$ be the production in the 1st year and $d$ be the fixed increase in production per year (common difference).
  The formula for the $n^{\text{th}}$ term of an AP is: $$a_n = a + (n - 1)d$$
  We are given:
      Production in the 3rd year ($a_3$) = $600$ $\implies a + 2d = 600$     --- (Equation 1)
      Production in the 7th year ($a_7$) = $700$ $\implies a + 6d = 700$     --- (Equation 2)
  
  Step 1: Find the production in the 1st year ($a$) and the annual increase ($d$):
  Subtract Equation 1 from Equation 2:
      $(a + 6d) - (a + 2d) = 700 - 600$
      $4d = 100$
      $d = \frac{100}{4} = 25$
  
  Substitute $d = 25$ into Equation 1:
      $a + 2(25) = 600$
      $a + 50 = 600$
      $a = 600 - 50 = 550$
      *(Thus, production in the 1st year is **$550$** sets).*
  
  Step 2: Find the production in the 10th year ($a_{10}$):
      $a_{10} = a + (10 - 1)d$
      $a_{10} = 550 + 9(25)$
      $a_{10} = 550 + 225 = 775$
      *(Thus, production in the 10th year is **$775$** sets).*
  
  Step 3: Find the total production in the first 7 years ($S_7$) using the formula $S_n = \frac{n}{2}[2a + (n - 1)d]$:
      $S_7 = \frac{7}{2}[2(550) + (7 - 1)25]$
      $S_7 = \frac{7}{2}[1100 + 6(25)]$
      $S_7 = \frac{7}{2}[1100 + 150]$
      $S_7 = \frac{7}{2}[1250]$
      $S_7 = 7 \times 625 = 4,375$
      *(Thus, total production in the first 7 years is **$4,375$** sets).*
  
  Result:
  (i) 1st year production = $550$ sets.
  (ii) 10th year production = $775$ sets.
  (iii) Total production in first 7 years = $4,375$ sets.

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