Class 10 Arithmetic Progressions Standard Worksheet

Question 1: Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.

Solution:
Step 1: $a_{11} = a + 10d = 38$ and $a_{16} = a + 15d = 73$.
Step 2: Subtracting eq(1) from eq(2): $5d = 35 \Rightarrow d = 7$.
Step 3: Substitute $d=7$ in eq(1): $a + 70 = 38 \Rightarrow a = -32$.
Step 4: $a_{31} = a + 30d = -32 + 30(7) = -32 + 210 = 178$.
Answer: 178.
Question 2: Which term of the AP: 3, 15, 27, 39, ... will be 132 more than its 54th term?

Solution:
Step 1: Here $a=3, d=12$. Let $a_n = a_{54} + 132$.
Step 2: $a + (n-1)d = a + 53d + 132 \Rightarrow (n-1)12 = 53(12) + 132$.
Step 3: Divide by 12: $n-1 = 53 + 11 \Rightarrow n-1 = 64 \Rightarrow n = 65$.
Answer: 65th term.
Question 3: Determine the AP whose 3rd term is 5 and the 7th term exceeds the 5th term by 10.

Solution:
Step 1: $a_7 - a_5 = 10 \Rightarrow (a+6d) - (a+4d) = 10 \Rightarrow 2d = 10 \Rightarrow d = 5$.
Step 2: $a_3 = a + 2d = 5 \Rightarrow a + 10 = 5 \Rightarrow a = -5$.
Answer: AP is -5, 0, 5, 10, ...
Question 4: Find the 20th term from the last term of the AP: 3, 8, 13, ..., 253.

Solution:
Step 1: Reverse the AP: 253, 248, ..., 13, 8, 3. New $a=253$, new $d = -5$.
Step 2: $a_{20} = a + 19d = 253 + 19(-5)$.
Step 3: $253 - 95 = 158$.
Answer: 158.
Question 5: 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 three terms of the AP.

Solution:
Step 1: $a_4 + a_8 = 24 \Rightarrow a+3d + a+7d = 24 \Rightarrow 2a+10d=24 \Rightarrow a+5d=12$.
Step 2: $a_6 + a_{10} = 44 \Rightarrow a+5d + a+9d = 44 \Rightarrow 2a+14d=44 \Rightarrow a+7d=22$.
Step 3: Subtracting: $2d=10 \Rightarrow d=5$. Then $a = 12 - 25 = -13$.
Answer: -13, -8, -3.
Question 6: Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year. In which year did his income reach ₹7000?

Solution:
Step 1: $a=5000, d=200, a_n=7000$.
Step 2: $7000 = 5000 + (n-1)200 \Rightarrow 2000 = (n-1)200 \Rightarrow 10 = n-1 \Rightarrow n=11$.
Step 3: 11th year means $1995 + 10 = 2005$.
Answer: In the year 2005.
Question 7: Find the sum of the first 40 positive integers divisible by 6.

Solution:
Step 1: AP is 6, 12, 18, ... Here $a=6, d=6, n=40$.
Step 2: $S_{40} = \frac{40}{2}[2(6) + 39(6)] = 20[12 + 234] = 20(246)$.
Step 3: $20 \times 246 = 4920$.
Answer: 4920.
Question 8: If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term.

Solution:
Step 1: $S_{14} = 1050, a=10$. Formula: $S_n = \frac{n}{2}[2a + (n-1)d]$.
Step 2: $1050 = \frac{14}{2}[20 + 13d] \Rightarrow 1050 = 7(20+13d) \Rightarrow 150 = 20+13d \Rightarrow 130 = 13d \Rightarrow d=10$.
Step 3: $a_{20} = a + 19d = 10 + 19(10) = 10 + 190 = 200$.
Answer: 200.
Question 9: The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Solution:
Step 1: $a=17, l=350, d=9$. $350 = 17 + (n-1)9 \Rightarrow 333 = 9(n-1) \Rightarrow 37 = n-1 \Rightarrow n=38$.
Step 2: $S_{38} = \frac{38}{2}(a+l) = 19(17+350) = 19(367)$.
Step 3: $19 \times 367 = 6973$.
Answer: $n=38$, Sum = 6973.
Question 10: A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ... . What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take $\pi = 22/7$)

Solution:
Step 1: Length of semicircle $l = \pi r$. Radii are 0.5, 1.0, 1.5...
Step 2: Lengths are $\pi(0.5), \pi(1.0), \pi(1.5)...$ This is an AP with $a=0.5\pi, d=0.5\pi$.
Step 3: Sum of 13 terms: $S_{13} = \frac{13}{2}[2(0.5\pi) + 12(0.5\pi)] = \frac{13}{2}[\pi + 6\pi] = \frac{13}{2}(7\pi)$.
Step 4: $\frac{91}{2} \times \frac{22}{7} = \frac{13 \times 11}{1} = 143$.
Answer: 143 cm.

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Chapter 5: Arithmetic Progressions

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