The Sum Of The Series 1 By 3 Into 7 Plus 1 By 7 Into 11 Is (1) 1/3 (2) 1/6 (3) 1/9 (4) 1/12 Solution: Let S = 1/3×7 + 1/7×11 + 1/11×15+... = (¼)(⅓ - 1/7) + ¼ (1/7 - 1/11) + ¼ (1/11 - 1/15)... View Article
The Sixth Hm Between 3 And 6 By 13 Is (1) 63/120 (2) 63/12 (3) 126/105 (4) 120/63 Solution: For HP, xn = (n+1)ab/(na+b) x6 = (7×3×6/13)/(6×3+6/13) = 126/240 = 63/120... View Article
The Fifth Term Of The Hp 2 2 And 1 By 2 3 And 1 By 3 Will Be (1) (2) (3) 1/10 (4) 10 Solution: 2, 2 ½ , 3 ⅓ are in HP. Ie, 2, 5/2, 10/3 are in HP. So ½, ⅖, 3/10 are in AP. Here a = ½... View Article
Which Number Should Be Added To The Numbers 13 15 19 So That The Resulting Numbers Be The Consecutive Terms Of A Hp (1) 7 (2) 6 (3) –6 (4) –7 Solution: Let x be the number added to 13, 15 and 19 . So 13+x, 15+x, 19+x are in HP. 2/(15+x) =... View Article
If A1 A2 A10 Be In Ap H1 H2 H10 Be In Hp A1 Equal H1 Eq 2 A10 Equal H10 Eq 3 Then A4h7 Equals (1) 2 (2) 3 (3) 5 (4) 6 Solution: Given a1, a2, ...a10 be in AP. a1 = h1 = 2, a10 = h10 = 3 a10 = a1+9d 3 = 2+9d 9d = 1 d =... View Article
The Sum Of Three Consecutive Terms In A Geometric Progression Is 14 If 1 Is Added To The First And The Second Terms And 1 Is Subtracted From The Third (1) 1 (2) 2 (3) 4 (4) 8 Solution: Let a, ar, ar2 be the three consecutive terms in GP. a+ar+ar2 = 14 a(1+r+r2) = 14..(i) a+1,... View Article
150 Workers Were Engaged To Finish A Piece Of Work In A Certain Number Of Days 4 Workers Dropped (1) 15 (2) 20 (3) 25 (4) 30 Solution: Let the number of days in which 150 workers finish the work be x. 150 x = 150+ 146 +... View Article
If The Ratio Of Hm And Gm Between Two Numbers A And B A Greater Than B Is 4 5 Then The Ratio Of The Two Numbers Will Be (1) 1 : 2 (2) 2 : 1 (3) 4 : 1 (4) 1 : 4 Solution: HM of a and b = 2ab/(a+b) GM of a and b = √(ab) Ratio of HM and GM = 4:5... View Article
Sum Of N Terms Of Series 12 16 24 40 Will Be (1) 2(2n-1)+8n (2) 2(2n-1)+6n (3) 3(2n-1)+8n (4) 4(2n-1)+8n Solution: Let S = 12 + 16 + 24 + 40 +...tn = (8+22)+ (8+23) +... View Article
If Three Positive Real Numbers A B C Are In Ap And If Abc Eq 64 Then The Minimum Value Of B Is (1) 6 (2) 5 (3) 4 (4) 3 Solution: We know AM ≥ GM (a+b+c)/3 ≥ (abc)1/3 …(i) Since a, b, c are in AP 2b = a+c So (i) becomes... View Article
If An Ap A Eq 1 Sn S2n Sn Eq Constant For All N Then The Common Difference D (1) 4 (2) 1/2 (3) 2 (4) 3 Solution: Given a = 1 S1/(S2-S1) = S2/(S4-S2) S1(S4-S2) = S2(S2-S1) S1S4-S1S2 = S22-S1S2 S22 =... View Article
The Nth Term Of An Ap Is P2 And The Sum Of The First N Terms Is S2 The First Term Is (1) (p2n+2s2)/n (2) (2s2+p2n)/n2 (3) (ps2-p2s)/n (4) (2s2-p2n)/n Solution: The nth term of AP, tn = a+(n-1)d = p2 Sum of first... View Article
If The Pth Term Of A Gp Is X And Q Th Term Is Y Then The Nth Term Is (1) (xn-p/yn-q)1/(p-q) (2) (xn+q/yn+p)1/(p-q) (3) (xn-q/yn-p)1/(p-q) (4) (xn-q/yn-q)1/(p+q) Solution: For a GP, pth term = arp-1 =... View Article
Sum Of Infinity Of The Series 1 Plus 4 5 7 By 5 2 Plus 10 By 5 3 Is (1) 7/16 (2) 5/16 (3) 104/64 (4) 35/16 Solution: Let S = 1+ ⅘ + 7/52 + 10/53 + …(i) S/5 = 1/5 + 4/52 + 7/53 + 10/53 +...∞... View Article
If A Is The Am Between A And B Then A 2b By A A Plus A 2a By A B (1) -8 (2) 2 (3) 4 (4) -4 Solution: Since A is the AM of a and b, A = (a+b)/2 (A-2b)/(A-a) = (a-3b)/(b-a) ..(i) (A-2a)/(A-b) =... View Article
In An Ap Sm Sn Eq M2 N2 The Ratio Of P2 Th Term To Q2 Term Is (1) (2p2+1)/(2q2+1) (2) (2p2-1)/(2q2-1) (3) (2p-1)/(2q-1) (4) (p2-2)/(q2-2) Solution: For an AP, S = (n/2)(2a+(n-1)d) Sm : Sn =... View Article
If 1 By A 1 By H 1 By B Are In Ap Then H Plus A By H A Plus H Plus B By H B (1) 2 (2) 4 (3) 0 (4) 1 Solution: Given 1/a, 1/H, 1/b are in AP. 2/H = 1/a + 1/b 2/H = (b+a)/ab H/2 = ab/(a+b) H = 2ab/(a+b)... View Article
If A1 A2 A10 Be In Ap 1 By H1 1 By H2 1 By H10 Be In Ap A1 Eq H1 Eq 2 A10 Eq H10 Eq 3 Then A4h7 (1) 1/6 (2) 6 (3) 3 (4) 2 Solution: Given a1, a2, ...a10 be in AP. a1 = h1 = 2, a10 = h10 = 3 a10 = a1+9d 3 = 2+9d 9d = 1 d =... View Article
If A 4 B Are In Ap And A 2 B Are In Gp Then 1 By A 1 1 By B Are In (1) AP (2) GP (3) HP (4) None of these Solution: Since a, 4, b are in AP 8 = a+b a, 2, b are in GP. So 4 = ab (a+b)/ab = 8/4... View Article
If For An Ap An A1 A5 A15 A26 A36 A40 Eq 210 Then S40 Eq (1) 2100 (2) 700 (3) 1400 (4) None of these Solution: a1 + a5 + a15 + a26 + a36 + a40 = 210 a+a+4d+a+14d+a+25d+a+35d+a+39d = 210... View Article