IMO Practice Test: Arithmetic Progressions | Class 10 Mathematics

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; \(a_{1}_{5}\)=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; \(S_{4}_{0}\)=3600; 40/2[2a+39d]=3600→20(2a+39d)=3600→2a+39d=180; After 30 installments, unpaid=3600/3=1200; \(S_{3}_{0}\)=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.

70386

71052

71928

72345

Explanation: Numbers: 101,108,115,...,997; AP with a=101,d=7; last term: 101+(n-1)7=997→(n-1)7=896→n-1=128→n=129; \(S_{1}_{2}_{9}\)=129/2(101+997)=129/2×1098=129×549=

Question 6 of 6 hard

Question 6 (see textbook)

Option A

Option B

Option C

Option D

Explanation:

IMO Practice Test — Arithmetic Progressions