Exercise 5.2 Practice
20 Practice Questions with Solutions
Q1: Table Problems
Fill in the blanks in the following table, given that $a$ is the first term, $d$ the common difference and $a_n$ the nth term of the AP:
(i) $a=3, d=2, n=10, a_n=\dots$
Formula: $a_n = a + (n-1)d$
$a_{10} = 3 + (10-1)(2)$
$a_{10} = 3 + 18 = 21$
Ans: 21
(ii) $a=-10, d=\dots, n=10, a_n=8$
$8 = -10 + (10-1)d$
$18 = 9d \Rightarrow d = 2$
Ans: 2
(iii) $a=\dots, d=-3, n=15, a_n=-5$
$-5 = a + (15-1)(-3)$
$-5 = a - 42 \Rightarrow a = 37$
Ans: 37
(iv) $a=-10, d=2.5, n=\dots, a_n=0$
$0 = -10 + (n-1)(2.5)$
$10 = 2.5(n-1)$
$4 = n-1 \Rightarrow n = 5$
Ans: 5
(v) $a=5, d=0, n=100, a_n=\dots$
Since $d=0$, all terms are same.
Ans: 5
Q2: MCQ Style
Choose the correct choice in the following:
(i) 20th term of the AP: 2, 5, 8... is?
$a=2, d=3, n=20$
$a_{20} = 2 + 19(3) = 2 + 57$
Ans: 59
(ii) 10th term of the AP: -2, 0, 2... is?
$a=-2, d=2, n=10$
$a_{10} = -2 + 9(2) = -2 + 18$
Ans: 16
Q3: Missing Terms
Find the missing terms in the boxes:
(i) 4, $\square$, 12
Middle term = Average of neighbors.
$x = \frac{4+12}{2} = 8$
Ans: 8
(ii) $\square$, 10, $\square$, 0
$a_2=10, a_4=0 \Rightarrow a+d=10, a+3d=0$
Subtract: $2d = -10 \Rightarrow d=-5$
$a = 10 - (-5) = 15$. 3rd term = $10-5=5$.
Ans: 15, 5
(iii) 2, $\square$, $\square$, 11
$a=2, a_4=11$.
$2 + 3d = 11 \Rightarrow 3d=9 \Rightarrow d=3$.
Ans: 5, 8
(iv) -3, $\square$, $\square$, $\square$, 5
$a=-3, a_5=5$.
$-3 + 4d = 5 \Rightarrow 4d=8 \Rightarrow d=2$.
Ans: -1, 1, 3
(v) $\square$, 20, $\square$, $\square$, $\square$, -20
$a_2=20, a_6=-20$.
$a+d=20, a+5d=-20$. Subtract: $4d=-40, d=-10$.
$a=30$.
Ans: 30, 10, 0, -10
Q4: Which Term?
Which term of the AP: 5, 10, 15, 20... is 105?
$a=5, d=5, a_n=105$.
$105 = 5 + (n-1)5$
$100 = 5(n-1) \Rightarrow 20 = n-1 \Rightarrow n=21$.
Ans: 21st term
Q5: Find Number of Terms
Find the number of terms in each of the following APs:
(i) 5, 10, ..., 100
$a=5, d=5, a_n=100$.
$100 = 5 + (n-1)5$.
$95 = 5(n-1) \Rightarrow 19 = n-1 \Rightarrow n=20$.
Ans: 20 terms
(ii) 20, 19.5, ..., -10
$a=20, d=-0.5, a_n=-10$.
$-10 = 20 + (n-1)(-0.5)$.
$-30 = (n-1)(-0.5) \Rightarrow 60 = n-1$.
Ans: 61 terms
Q6: Check Term
Check whether 50 is a term of the AP: 2, 6, 10...
$a=2, d=4$. Let $a_n = 50$.
$50 = 2 + (n-1)4 \Rightarrow 48 = 4(n-1)$.
$12 = n-1 \Rightarrow n=13$.
Yes, it is the 13th term.
Q7: Two Conditions
Find the 31st term of an AP whose 10th term is 20 and the 20th term is 40.
$a+9d=20$, $a+19d=40$.
Subtract: $10d=20 \Rightarrow d=2$.
$a + 18 = 20 \Rightarrow a=2$.
$a_{31} = 2 + 30(2) = 62$.
Ans: 62
Q8: Finite AP
An AP consists of 50 terms. The 3rd term is 5 and the last term is 99. Find the 29th term.
$a_3=5 \Rightarrow a+2d=5$.
$a_{50}=99 \Rightarrow a+49d=99$.
Subtract: $47d = 94 \Rightarrow d=2$.
$a + 4 = 5 \Rightarrow a=1$.
$a_{29} = 1 + 28(2) = 57$.
Ans: 57
Q9: Find Zero Term
If the 3rd term is 4 and 9th term is -8, which term is zero?
$a+2d=4$, $a+8d=-8$.
$6d = -12 \Rightarrow d=-2$. So $a=8$.
Let $a_n=0 \Rightarrow 8 + (n-1)(-2) = 0$.
$8 = 2(n-1) \Rightarrow 4=n-1$.
Ans: 5th term
Q10: Exceeding Term
The 15th term of an AP exceeds its 10th term by 25. Find the common difference.
$a_{15} - a_{10} = 25$.
$(a+14d) - (a+9d) = 25$.
$5d = 25 \Rightarrow d=5$.
Ans: 5
Q11: Complex Condition
Which term of the AP: 3, 15, 27, 39... will be 120 more than its 20th term?
$d = 12$. Find $n$ such that $a_n = a_{20} + 120$.
$a + (n-1)d = a + 19d + 120$.
$(n-1)12 = 19(12) + 120$. Divide by 12.
$n-1 = 19 + 10 = 29 \Rightarrow n=30$.
Ans: 30th term
Q12: Difference of Terms
Two APs have the same common difference. The difference between their 100th terms is 100. What is the difference between their 1000th terms?
Let APs be $A_n$ and $B_n$.
$A_{100} - B_{100} = (a + 99d) - (b + 99d) = a - b = 100$.
Diff between 1000th terms: $(a + 999d) - (b + 999d) = a - b$.
Since $a-b=100$, the diff is always 100.
Ans: 100
Q13: Divisibility
How many three-digit numbers are divisible by 6?
First term: 102. Last term: 996. $d=6$.
$996 = 102 + (n-1)6$.
$894 = 6(n-1) \Rightarrow 149 = n-1$.
Ans: 150 numbers
Q14: Multiples
How many multiples of 4 lie between 12 and 248?
AP: 12, 16, ..., 248. (Both included).
$248 = 12 + (n-1)4$.
$236 = 4(n-1) \Rightarrow 59 = n-1$.
Ans: 60 terms
Q15: Equal Terms
For what value of n, are the nth terms of two APs: 10, 12, 14... and 5, 8, 11... equal?
AP1: $10 + (n-1)2$. AP2: $5 + (n-1)3$.
$10 + 2n - 2 = 5 + 3n - 3$.
$8 + 2n = 2 + 3n \Rightarrow n = 6$.
Ans: 6th term
Q16: Determine AP
Determine the AP whose 3rd term is 5 and the 7th term exceeds the 5th term by 10.
$a_7 - a_5 = 10 \Rightarrow 2d = 10 \Rightarrow d=5$.
$a_3 = 5 \Rightarrow a + 10 = 5 \Rightarrow a = -5$.
Ans: -5, 0, 5, 10...
Q17: From the End
Find the 10th term from the last term of the AP: 2, 4, 6, ..., 100.
Reverse AP: $100, 98, ..., 2$. $a=100, d=-2$.
$a_{10} = 100 + 9(-2) = 100 - 18$.
Ans: 82
Q18: Sum Equations
The sum of the 3rd and 7th terms of an AP is 20 and the sum of the 5th and 9th terms is 32. Find the AP.
Eq1: $(a+2d) + (a+6d) = 20 \Rightarrow 2a+8d=20$.
Eq2: $(a+4d) + (a+8d) = 32 \Rightarrow 2a+12d=32$.
Subtract: $4d=12 \Rightarrow d=3$.
$2a + 24 = 20 \Rightarrow 2a = -4 \Rightarrow a=-2$.
Ans: -2, 1, 4...
Q19: Word Problem (Salary)
A person started work at an annual salary of ₹10,000 with an increment of ₹500 each year. In which year did his income reach ₹20,000?
$a=10000, d=500, a_n=20000$.
$20000 = 10000 + (n-1)500$.
$10000 = 500(n-1) \Rightarrow 20 = n-1 \Rightarrow n=21$.
Ans: 21st year
Q20: Word Problem (Savings)
Priya saved ₹10 in the first week and increased her savings by ₹5 weekly. If her savings become ₹110 in the nth week, find n.
$a=10, d=5, a_n=110$.
$110 = 10 + (n-1)5$.
$100 = 5(n-1) \Rightarrow 20 = n-1$.
Ans: n = 21