Class 10 Arithmetic Progression Mock Test (ICSE)

Topics in Arithmetic Progression

• General term of an AP
• Sum of first n terms
• Finding an AP from given conditions
• Word problems on AP

Arithmetic Progression practice questions with answers

1. Q1Multiple choice·1 mark
   Find the sum of the first $15$ terms of the AP: $3, 7, 11, 15, \dots$

   1. (a)$465$
   2. (b)$480$
   3. (c)$450$
   4. (d)$495$
   Show answer and step marks

   1. Correct option: (a) $465$ (using $S_n = \frac{n}{2}[2a + (n-1)d]$ with $a=3, d=4, n=15$)+1

   Answer: (a)
2. Q2Multiple choice·1 mark
   If the $3^{\text{rd}}$ term of an AP is $5$ and the $7^{\text{th}}$ term is $9$, then its first term is:

   1. (a)$1$
   2. (b)$2$
   3. (c)$3$
   4. (d)$5$
   Show answer and step marks

   1. Correct option: (c) $3$ (solving $a + 2d = 5$ and $a + 6d = 9$ gives $d = 1$ and $a = 3$)+1

   Answer: (c)
3. Q3Multiple choice·1 mark
   Which term of the AP: $21, 18, 15, \dots$ is $0$?

   1. (a)$10^{\text{th}}$ term
   2. (b)$8^{\text{th}}$ term
   3. (c)$7^{\text{th}}$ term
   4. (d)$9^{\text{th}}$ term
   Show answer and step marks

   1. Correct option: (b) $8^{\text{th}}$ term (using $21 + (n-1)(-3) = 0 \implies n = 8$)+1

   Answer: (b)
4. Q4Multiple choice·1 mark
   A man saves ₹ $100$ in the first month and increases his saving by ₹ $50$ every month. In which month will his monthly saving be ₹ $550$?

   1. (a)$9^{\text{th}}$ month
   2. (b)$10^{\text{th}}$ month
   3. (c)$11^{\text{th}}$ month
   4. (d)$12^{\text{th}}$ month
   Show answer and step marks

   1. Correct option: (b) $10^{\text{th}}$ month (using $a_n = a + (n-1)d \implies 100 + (n-1)50 = 550 \implies n = 10$)+1

   Answer: (b)
5. Q5Long answer·4 marks
   The sum of the $4^{\text{th}}$ and $8^{\text{th}}$ terms of an Arithmetic Progression 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. Hence, write the first four terms of the AP and find its $20^{\text{th}}$ term.
   Show answer and step marks

   1. Using $a_n = a + (n-1)d$: $(a + 3d) + (a + 7d) = 24 \implies 2a + 10d = 24 \implies a + 5d = 12$.+0.5
   2. $(a + 5d) + (a + 9d) = 44 \implies 2a + 14d = 44 \implies a + 7d = 22$.+0.5
   3. Subtract the first equation from the second: $2d = 10 \implies d = 5$.+0.5
   4. Substitute $d = 5$ in $a + 5d = 12$: $a = 12 - 25 = -13$.+0.5
   5. Write the first four terms of the AP: $-13, -8, -3, 2$.+1
   6. Calculate the $20^{\text{th}}$ term: $a_{20} = a + 19d = -13 + 19(5) = -13 + 95 = 82$.+1

   Answer: First term $a = -13$, common difference $d = 5$, AP: $-13, -8, -3, 2, \dots$, $20^{\text{th}}$ term $= 82$
6. Q6Long answer·4 marks
   The sum of the 4th and 8th terms of an AP is $24$ and the sum of the 6th and 10th terms is $44$. Find the first term, the common difference, and the first three terms of the AP.
   Show answer and step marks

   1. Using the nth term formula $a_n = a + (n-1)d$, express the 4th and 8th terms and set up the first equation: $(a + 3d) + (a + 7d) = 24 \implies 2a + 10d = 24$, which simplifies to $a + 5d = 12$.+1
   2. Express the 6th and 10th terms and set up the second equation: $(a + 5d) + (a + 9d) = 44 \implies 2a + 14d = 44$, which simplifies to $a + 7d = 22$.+1
   3. Subtract the first simplified equation from the second: $(a + 7d) - (a + 5d) = 22 - 12 \implies 2d = 10 \implies d = 5$.+0.5
   4. Substitute $d = 5$ into $a + 5d = 12$: $a = 12 - 25 = -13$.+0.5
   5. Calculate the first three terms: $a_1 = -13$, $a_2 = -13 + 5 = -8$, $a_3 = -13 + 10 = -3$. State the final AP: $-13, -8, -3, \dots$+1

   Answer: First term $a = -13$, common difference $d = 5$, and the AP is $-13, -8, -3, \dots$

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