If The Angles A B C Of A Triangle Abc Are In Ap Then 1) c2 = a2 + b2 – ab2 2) c2 = a2 + b2 3) b2 = a2 + c2 – ac2 4) a2 = b2 + c2 – bc Solution: Let the angles be a-d, a, a+d. Adding... View Article
If The Numbers P Q R Are In Ap Then 2 P2 2 Pq 2 Pr Are In (1) A.P. (2) G.P. (3) H.P. (4) A.G.P. Solution: Given p,q,r are in AP. 2q = p+r (2p)2q = (2p)p+r 22pq = .2pr (2pq)2 = .2pr So , 2pq,... View Article
3 By 1 2 Plus 5 By 1 2 Plus 2 2 Plus 7 By 1 2 Plus 2 2 Plus 32 N Terms Equal (1) 6n2/(n+1) (2) 6n/(n+1) (3) 6(2n-1)/(n+1) (4) 3(n2+1)/(n+1) Solution: Given series 3/12 + 5/12+22 + 7/(12+22+32) +...n terms... View Article
The Nth Term Of The Sequence 1 By 1 Plus Root X 1 By 1 X 1 By 1 Root X Is (1) (1+√x(n2-2))/(1-x) (2) (1+√x(n-1))/(1+√x) (3) (1+√x(n-2))/(1-x) (4) (3-√x(n+2))/3(1-x) Solution: Given series 1/1+√x, 1/1-x,... View Article
If The Sides Of A Right Triangle Are In Ap Then The Sum Of The Sines Of The Two Acute Angles Is (1) 7/5 (2) 8/5 (3) 1/5 (4) 6/5 Solution: Let the sides of the triangle be a-d, a, a+d. (a+d)2 = a2+(a-d)2 a2+2ad+d2 =... View Article
The Numbers Of Terms In The Ap A B C X Is (1) (x+b+a)/(c-a) (2) (x+b-2a)/(c-b) (3) (x+b+2a)/(c-b) (4) (x-b+a)/(c-b) Solution: Given a,b,c...x are in AP Common difference =... View Article
The Sum Of The Series A A Plus D Plus A Plus 2d Up To 50 Terms Is (1) -50d (2) 25d (3) a + 50d (4) -25d Solution: a - (a + d) + (a + 2d) - (a + 3d) + ... up to 50 terms = -d+-d+...25 terms = -25d... View Article
Sum Of Products Of First N Natural Numbers Taken Two At A Time Is (1) n(n2-1)(3n+2)/24 (2) n(n+1)2(3n+2)/74 (3) n2(n+1)(3n+2)/48 (4) n(n+1)(n+2)(3n+2)/96 Solution: We know (x1 + x2 + ...xn)2 = x12... View Article
If The First Second And Last Terms Of An Ap Are A B And 2a Respectively The Sum Of The Series Is (1) ab/2(b-a) (2) 3ab/2(b-a) (3) ab/(b-a) (4) none of these Solution: Given first term = a Second term = b Last term = 2a... View Article
If An Is An Ap Then A12 A22 A32 A42 A992 A1002 Equal (1) (50/99)(a12-a1002) (2) (100/99)(a1002-a12) (3) (50/51)(a12-a1002) (4) none of these Solution: Given a1, a2, a3….an is in AP.... View Article
If The Sum Of The Roots Of The Equation Ax2 Plus Bx Plus C Equal 0 Is Equal To The Sum Of The Squares Of Their Reciprocals Then Bc2 Ca2 Ab2 Are In (1) AP (2) GP (3) HP (4) AGP Solution: Let p and q be the roots of the equation ax2 + bx + c = 0 Then p+q = -b/a pq = c/a Given... View Article
If Sn Equal An Plus Bn2 For An Ap Where A And B Are Constants Then Common Difference Of Ap Will Be (1) 2b (2) a + b (3) 2a (4) a - b Solution: Given Sn = an+bn2 S1 = a+b So first term = a+b S2 = 2a+4b = 2(a+2b) S2-S1 =... View Article
If A Set A Equal 3 7 11 407 And A Set B Equal 2 9 16 709 Then N A Intersection B Equal (1) 13 (2) 14 (3) 15 (4) 16 Solution: Given A = {3, 7, 11, ..., 407} and B = {2, 9, 16, ..., 709} For A, common difference, d = 4... View Article
If P Th Term And Q Th Term Of An Ap Are 1 By Qr And 1 By Pr Respectively Then R Th Term Of The Ap Equal (1) 1/pqr (2) 1 (3) 1/pq (4) pq Solution: Given pth term = 1/qr qth term = 1/pr a+(p-1)d = 1/qr a+(q-1)d = 1/pr (p-q)d = 1/qr... View Article
If An Ap T35 Equal 50 And D Equal 3 Then S35 Equal (1) 35 (2) 38 (3) 32 (4) 29 Solution: Given T35 = -50, d = -3 a+34d = -50 a+34×-3 = -50 a-102 = -50 a = -50+102 a = 52 Sn =... View Article
The Series 1 1 Fact Plus 2 2fact Plus 3 3 Fact Plus N N Fact Equal (1) (n + 1)! -n (2) (n + 1)! -1 (3) n! -1 + n (4) n! +1 -n Solution: 1. 1! + 2. 2! + 3. 3! + ... + n. n! = ∑n= 1n n.n! = ∑n=... View Article
First Term Of A Gp Of 2n Terms Is A And The Last Term Is L The Product Of All The Terms Of The Gp Is (1) (al)n/2 (2) (al)n-1 (3) (al)n (4) (al)2n Solution: Given first term = a Last term = ar2n-1 = l Number of terms = 2n Product... View Article
The Sum Of The Numbers 1 Plus 2 2 Plus 3 22 Plus 4 23 Plus 50 249 Is (1) 1 + 49. 249 (2) 1 + 49. 250 (3) 1 + 50. 249 (4) 1 + 50. 250 Solution: Let S = 1 + 2.2 + 3. 22 + 4. 23 + ... + 50. 249…(i) 2S... View Article
In A Gp The Last Term Is 1024 And The Common Ratio Is 2 Its 20 Th Term From The End Is (1) 1/512 (2) 1/1024 (3) 1/256 (4) 512 Solution: Given r = 2 arn-1 = 1024 a, ar, ar2 ….1024 is a GP with r = 2. When... View Article
If 1 A1 A2 A3 An 31 Are In Ap And A7 An 1 Equal 5 By 9 Then N Equal (1) 28 (2) 14 (3) 15 (4) 13 Solution: Given 1, A1, A2, A3, .... An, 31 are in A. P. So 1+(n+1)d = 31 d = 30/(n+1)...(i) A7 : An... View Article