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1) How many numbers between 15 & and 190 are divisible by 11?A) 9 B) 8 C) 15 D) 16
Answer & Explanation
Answer: 16 (The required numbers are 22,33,44,55 · · ·187
This is an A.P with a=22 and d = (33-22)=11
let it contain n terms.
Tn=187 =>a+(n-1)d=187
22+(n-1)*11=187
n=16)
2) The sum of first 3 numbers of a G.P is 39.The sum of first & 3rd no is 30.What is the 8th term of the progressionA) 2187 B) 6561 C) 729 D) 19683
Answer & Explanation
Answer: 6561 (Let a,b,c be the first three terms of the progression. Given a+b+C = 39 & a+c = 30; b =9.In a G.P b2 = a*c & b = 9 & the progressions is 3,9,27
an = a1 *r n-1,a≠0.r≠0 Where an is the nth term a1 is the first term and r is the common ratio.
8th term of the progressions is 6561)
3) Three numbers a,b,12 form an A.P,Three numbers a,b,16 form a G.P.Find a & b.A) 4,8 or 36,24 B) 36,24 C) 4,8 D) 8,4 or 24,36
Answer & Explanation
Answer: 4,8 or 36,24 (In the G.P b2 = 16a. In the a.p b-a = 12-b; b = (12+a)/2 ;16a =[(12+a)/2]2; solving this ,a = 4 or 36 & b= 24 or 8)
4) The Sum of 1st & 5th term of an A.P is 36 & the product of 2nd & 4th term is 308.What is the sum of first & 8th term?A) 46 B) 48 C) 56 D) 52
Answer & Explanation
Answer: 48 (Let the A.P be (a-2d),(a-d),a(a+d),(a+2d)
(a-2d)+(a+2d) = 36 a= 18
(a+d)(a-d) = a2 - d2=308
d2=324-308 = 16 & d= 4.
Sum of 1st & 8th term is 48.)
5) Consider a sequence of 10 consecutive numbers.If arithmetic mean of first 5 integers is n , what is the arithmetic mean first 9 integersA) n+1 B) n+(1/2) C) n+3 D) n+2
Answer & Explanation
Answer: n+2 (Shortcut: Assume the 10 consecutive numbers are 1,2, ···10. Mean of first 5 digits is (1+2+3+4+5)/5 = 3;
Mean of first 9 digits is (1+2+····+9)/9 = 5.Answer is (n+2))
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