The Value Of 1 By 1 Plus A By 2 Plus A Plus 1 By 2 Plus A By 3 Plus A Infinity Is Where A Is A Constant (1) 1/(1+a) (2) 2/(1+a) (3) ∞ (4) None of these Solution: 1/(1+a)(2+a) + 1/(2+a)(3+a) + 1/(3+a)(4+a) +..... = 1/(1+a) - 1/(2+a) +... View Article
If 1 By 1 4 Plus 1 By 2 4 Plus 1 By 3 4 Infinity Pi 4 90 Then The Value Of 1 By 1 4 Plus 1 By 3 4 Plus 1 By 5 4 Infinity Is (1) π4/96 (2) π4/45 (3) 89 π4/90 (4) none of these Solution: Given 1/14 + 1/24 + 1/34 + 1/44 + ...∞ = π4/90 (1/14 + 1/34 + 1/54... View Article
If Tn Eq 1 By 4 N Plus 2 N Plus 3 For N Eq 1 2 3 Then 1 T1 1 T2 1 T3 1 T2003 (1) 4006/3006 (2) 4003/3007 (3) 4006/3008 (4) 4006/3009 Solution: Given tn = (1/4)(n+2)(n+3) t1 = (1/4)(3×4) t2 = (¼)×4×5 t3 =... View Article
Four Numbers Are In Arithmetic Progression The Sum Of The First And Last Term Is 8 And The Product Of Both Middle Terms Is 15 The Least Number Of The Series Is (1) 4 (2) 3 (3) 2 (4) 1 Solution: Let a-3d, a-d , a+d, a+3d are in AP. The sum of the first and last term is 8. a-3d+a+3d = 8... View Article
If A B C D E Are In Ap Then The Value Of A B 4c 4d E In Terms Of A If Possible Is (1) 4a (2) 2a (3) 3 (4) None of these Solution: Given a,b,c,d,e are in A.P. Let D be the common difference of AP. b = a+D c =... View Article
If 1 Log9 3 1 X Plus2 Log3 4 3x 1 Are In Ap Then X Equals (1) log3 4 (2) 1 – log3 4 (3) 1 – log4 3 (4) log4 3 Solution: Given 1, log9 (31– x + 2), log3(4.3x-1) are in A.P So 2log9 (31– x... View Article
Let Tr Be The Rth Term Of An Ap For R 1 2 3 If For Some Positive Integers M N We Have Tm Eq 1 By N And Tn Eq 1 By M Then Tmn Equal (1) 1/mn (2) 1/m + 1/n (3) 1 (4) 0 Solution: Let a be the first term and d be the common difference of AP. rth term = Tr Tm... View Article
If Am Denotes The Mth Term Of An Ap Then Am (1) 2/(am+k + am-k) (2) (am+k - am-k)/2 (3) (am+k + am-k)/2 (4) None of these Solution: Let a be the first term and d be the... View Article
If The Pth Qth And Rth Term Of An Arithmetic Sequence Are A B And C Respectively Then The Value Of (1) 1 (2) –1 (3) 0 (4) 1/2 Solution: Let A be the first term and D be the common difference. pth term = A+(p-1)D = a a = A-D+pD... View Article
The 9th Term Of The Series 27 Plus 9 Plus 5 And 2 By 5 Plus 3 And 6 By 7 Will Be (1) (2) 10/17 (3) 16/27 (4) 17/27 Solution: = 27 + 27/3 + 27/5 + 27/7 +....27/(2n-1) Tn = 27/(2n-1) T9 = 27/(18-1) = 27/17... View Article
The First And Last Terms Of A Gp Are A And L Respectively R Being Its Common Ratio Then The Number Of Terms In This Gp Is (1) (log l - log a)/ log r (2) 1 - (log l - log a)/ log r (3) (log a - log l)/ log r (4) 1 + (log l - log a)/ log r Solution:... View Article
Pth Term Of The Series 3 1 By N Plus 3 2 By N Plus 3 3 By N Will Be (1) 3 + p/n (2) 3 - p/n (3) 3 + n/p (4) 3 - n/p Solution: Here first term, a = 3 - 1/n Common difference, d = -1/n Pth term, ap... View Article
If Arithmetic Mean Of Two Positive Numbers Is A Their Geometric Mean Is G And Harmonic Mean Is H Then H Is Equal To (1) G2/A (2) G/A2 (3) A2/G2 (4) A/G2 Solution: Let a and b be two numbers. AM, A = (a+b)/2 GM, G = √(ab) HM, H = 2ab/(a+b) =... View Article
If A1 A2 An Are Positive Real Numbers Whose Product Is A Fixed Number C Then The Minimum Value Of A1 A2 An 1 2an Is (1) n(2c)1/n (2) (n + 1)c1/n (3) 2nc1/n (4) (n + 1)(2c)1/n Solution: Given product, a1a2...an = c a1a2…(an-1)(2an) = 2c ….(i) We... View Article
If A B C Are In Ap And A 2 B 2 C 2 Are In Hp Then (1) a ≠b ≠c (2) a2 = b2 = c2/2 (3) a, b, c are in G.P (4) -a/2, b, c are in GP Solution: Since a, b, c are in AP 2b = a+c …(i)... View Article
If A By B B By C C By A Are In Hp Then (1) a2b, c2a, b2c are in AP (2) a2b, b2c, c2a are in HP (3) a2b, b2c, c2a are in GP (4) none of these Solution: since a/b, b/c,... View Article
If A B C Are In Ap Then 2ax Plus 1 2bx Plus 1 2cx Plus 1 X Not Equal 0 Are In (1) A.P (2) G.P only when x > 0 (3) G.P if x < 0 (4) G.P for all x ≠0 Solution: Since a, b, c are in AP 2b = a+c (2bx... View Article
If In The Equation Ax2 Bx C Eq 0 The Sum Of The Roots Is Equal To The Sum Of The Squares Of Their Reciprocals Then C By A A By B B By C Are In (1) AP (2) GP (3) HP (4) none of these Solution: Let p and q be the roots of the equation ax2 + bx + c = 0 Then sum of roots, p+q... View Article
If A1 A2 G1 G2 And H1 H2 Be Two Ams Gms And Hms Between Two Numbers Respectively Then The Value Ofg1g2 By H1h2 H1 Plus H2 By A1 Plus A2 (1) 1 (2) 0 (3) 2 (4) 3 Solution: Let a, A1, A2, b are in AP. Then (a+b)/2 = (A1+A2)/2 (a+b) = (A1+A2) ..(i) Let a, G1, G2, b... View Article
If First Three Terms Of Sequence 1 By 16 A B 1 By 6 Are In Geometric Series And Last Three Terms Are In Harmonic Series (1) a = -1/4, b = 1 (2) a = 1/12, b = 1/9 (3) (1) and (2) both are true (4) none of these Solution: Given 1/16, a, b are in GP.... View Article
