IMO Practice Test — Arithmetic Progressions
6 Questions • 15 min • Olympiad level
15:00
Question 1 of 6
medium
If the sum of first n terms of an AP is \(3n^{2} + 5n\), find its 15th term.
86
92
98
104
Explanation: S_n=\(3n^{2}+5n\); a_n=S_n-S_{n-1}=[\(3n^{2}+5n\)]-[3(n-1)\(^{2}+5\)(n-1)] = 6n+2; \(a_{15}\)=6×15+2=92
Question 2 of 6
medium
Three numbers in AP have sum 30 and product 910. Find the largest number.
10
13
15
17
Explanation: Let numbers: a-d,a,a+d; sum=3a=30→a=10; product=10(100-\(d^{2}\))=910→100-\(d^{2}\)=91→\(d^{2}\)=9→d=±3; numbers:7,10,13 or 13,10,7; largest=13
Question 3 of 6
medium
A man arranges to pay off a debt of ₹3600 in 40 annual installments which form an AP. When 30 installments are paid, he dies leaving one-third of the debt unpaid. Find the first installment.
₹40
₹45
₹51
₹55
Explanation: Total debt=3600; \(S_{40}\)=3600; 40/2[2a+39d]=3600→20(2a+39d)=3600→2a+39d=180; After 30 installments, unpaid=3600/3=1200; \(S_{30}\)=2400; 30/2[2a+29d]=2400→15(2a+29d)=2400→2a+29d=160; Subtract: 10d=20→d=2; then 2a+78=180→2a=102→a=51
Question 4 of 6
medium
If the sum of m terms of an AP is equal to the sum of n terms, prove that sum of (m+n) terms is zero. Which of the following conditions must be true?
a = d
a + d = 0
2a + (m+n-1)d = 0
a = -d
Explanation: S_m=S_n → m/2[2a+(m-1)d]=n/2[2a+(n-1)d]; rearrange: (m-n)2a + (\(m^{2}-m-n^{2}+n\))d=0; (m-n)[2a+(m+n-1)d]=0; since m≠n, 2a+(m+n-1)d=0; then S_{m+n}=(m+n)/2[2a+(m+n-1)d]=0
Question 5 of 6
medium
A sequence is formed by all 3-digit numbers that leave remainder 3 when divided by 7. Find the sum of all numbers in this sequence.
70281
70821
71928
72345
Explanation: The numbers are $101, 108, 115, \dots, 997$, an AP with $a=101,\ d=7$. From $997=101+(n-1)7$ we get $n=129$. $S_{129}=\frac{129}{2}(101+997)=\frac{129}{2} \times 1098 = 129 \times 549 = 70821$.
Question 6 of 6
hard
In an AP, the 4th term is zero. Find the ratio of its 25th term to its 11th term.
1:2
2:1
3:1
4:1
Explanation: Given the 4th term is $0$: $a+3d=0$, so $a=-3d$. Then $a_{11}=a+10d=7d$ and $a_{25}=a+24d=21d$. Hence $a_{25}:a_{11}=21d:7d=3:1$.