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If A1 A2 An Are In Ap With Common Difference D Not Equal 0 Then Sin D Sec A Is Equal To

(1) cot an + cot a1 (2) cot an - cot a1 (3) tan an - tan a1 (4) tan an - tan an-1 Solution: Given a1, a2….an are in AP.... View Article

If The First Second And Last Terms Of An Arithmetic Series Are A B And C Respectively Then The Number Of Terms Is

Solution: First term = a Second term = b Common difference, d = b-a Last term, (a+(n-1)d = c a+(n-1)(b-a) = c (n-1)(b-a) = (c-a) (n-1) =... View Article

If 5 9 13 N Terms 7 9 11 12 Terms 5 12 Then N Equals

(1) 5 (2) 6 (3) 9 (4) 12 Solution: Let S1 = 5+9+13+....n terms Here a = 5, d = 4 S1 = (n/2)(2a+(n-1)d) = (n/2)(10+(n-1)4) =... View Article

If The Sum Of First N Terms Of An Ap Is Cn2 Then The Sum Of Squares Of These N Terms Is

(1) n(4n2-1)c2/6 (2) n(4n2+1)c2/3 (3) n(4n2-1)c2/3 (4) none of these Solution: Given Sn = cn2 Sn-1 = c(n-1)2 = cn2+c-2cn Tn... View Article

An Ap Consists Of 23 Terms If The Sum Of The Three Terms In The Middle Is 141 And The Sum Of The Last Three Terms Is 261 Then The First Term Is

(1) 6 (2) 5 (3) 4 (4) 3 Solution: Let a1,a2,...a23 be the terms in AP. Let d be the common difference. Given a11+a12+a13 =... View Article

The Sum Of All Two Digit Natural Numbers Which Leave A Remainder 5 When They Are Divided By 7 Is Equal To

(1) 715 (2) 702 (3) 615 (4) 602 Solution: The numbers are 12, 19, ....96 Here a = 12, d = 19-12 = 7 Last term, L = 96... View Article

The Arithmetic Mean Of 7 Consecutive Integers Starting With A Is M Then The Arithmetic Mean Of 11 Consecutive Integers Starting With A 2 Is

(1) 2a (2) 2m (3) a+4 (4) m+4 Solution: Given the arithmetic mean of 7 consecutive integers starting with a is m. So common difference =... View Article

If The Sum To First N Terms Of An Ap 2 4 6 Is 240 Then The Value Of N Is

(1) 14 (2) 15 (3) 16 (4) 17 Solution: Here first term a = 2 Common difference, d = 4-2 = 2 Sum of n terms = (n/2)(2a+(n-1)d) =... View Article

A Person Is To Count 4500 Currency Notes Let An Denote The Number Of Notes He Counts In The Nth Minute

(1) 24 min (2) 34 min (3) 125 min (4) 135 min Solution: There are a total 4500 notes. Given a1 = a2 = ... = a10 = 150 So the... View Article

If S1 S2 And S3 Are The Sum Of N 2n And 3n Terms Respectively Of An Arithmetic Progression Then

(1) S3 = 2(S1 + S2 ) (2) S3 = S1 + S2 (3) S3 = 3(S2 - S1) (4) S3 = 3(S2 + S1) Solution: S1 = (n/2)(2a+(n-1)d) …(i) S2 =... View Article

The First Four Terms Of An Ap Are A 9 3a B 3a B The 2011 Th Term Of An Ap Is

(1) 2015 (2) 4025 (3) 5030 (4) 8045 Solution: Given a, 9, 3a-b, 3a+b are in AP. Common difference d = 9-a Second term - first... View Article

If The Numbers A B C D And E Are In The Form Of An Ap With A 1 Then A 4b 6c 4d E Is Equal To

(1) 1 (2) 2 (3) 0 (4) 3 Solution: Given a,b,c,d,e are in AP. Let D be the common difference. b = a+D c = a+2D d = a+3D e =... View Article

Let An Be The Nth Term Of An Ap If Sigma R 1 To 100 Alpha And A2r 1 Beta Then The Common Difference Of An Ap Is

(1) (α-β)/200 (2) (α-β)/100 (3) (α-β) (4) β-α Solution: Given ∑r=1100 a2r = α ∑r=1100 a2r-1 = β a2+a4+a6+....+a200 = α …(i)... View Article

Six Numbers Are In An Ap Such That Their Sum Is 3 The First Term Is 4 Times The Third Term Then The Fifth Term Is

(1) - 15 (2) - 3 (3) 9 (4) - 4 Solution: Let a-5d, a-3d, a-d, a+d, a+3d, a+5d are the numbers in AP. Sum = 3 a-5d+ a-3d+... View Article

If 100 Times The 100th Term Of An Ap With Non Zero Common Difference Equals To 50 Times Its 50th Term Then The 150th Term Of An Ap Is

(1) - 150 (2) 150 times its 50th term (3) 150 (4) Zero Solution: Let d be the common difference and a be the first term of an... View Article