If a,b,c,d are in G.P. while a-2,b-7,c-9,d-5 are in A.P., then a+b+c+d is divided by?

a)2 b)3 c)5 d)7

2 years ago


Answers : (1)


2 years ago

Post Your Answer

More Questions On Algebra

Ask Experts

Have any Question? Ask Experts
Post Question
Answer ‘n’ Earn
Attractive Gift
To Win!!!
Click Here for details
the gcd of two positive integer is 81 and their lcm is 5103 , find the number
i want answer from you
anand kapare 2 months ago
729 and 567 GCD=81 Therefore numbers are 81a and 81b for a and b are co-prime. LCM=81*a*b = 5103=81*63 therefore ab=63 a,b=9,7 so numbers are 81*7=567 and 81*9 =729 GCD of 15,20,35 is 5...
Karan Yadav 2 months ago
find gcd of 15,20,35 . also find x, y, z such that 15x+20y+35z= gcd
anand kapare 2 months ago
A cricket player played n (n>1) matches during his career and made a total of (n^2-12n+39)(4.6^n-5.3^n+1)/5 runs. If T r represents the runs made by the player in r^th match such that T 1...
Hello student, Please find my response to your question below I think there is some information part missing in the above question so iam not able to solve this problem please recheck the...
SHAIK AASIF AHAMED one month ago
If AB=2a, given base of triangle ABP where p is a varianle such that tanA+tanB=k(constant) then locus of P is? a: circle b:parabola c: hyperbola
b:parabola is ans. use A+B = 180-P then take tan and apply angle between lines formula. Assume A(a,0) B(-a,0)
yashpal 16 days ago
3 ∫ 2 (2x 2 + 1) dx
integration of (2(X^2)+1)dx from 2 to 3 is equal to (2((x^3)/3)+x) from 2 to3 =(2((3^3)/3)+3)-(2((2^3)/3)+2) =(2*(27/3)+3)-(2*(8/3)+2) =(18+3)-((16/3)+2) =(21)-(22/3) =41/3
Miryala Gopalakrishna 27 days ago
Ans:- 3 ∫ 2 (2x 2 +1)dx =2[(x 2+1 /2+1)] 2 3 +[x] 2 3 =2[(x 3 /3)] 2 3 +[x] 2 3 =2[(3 3 /3)+(2 3 /3)]+[3-2] =2[(27+8)/3]+1 =2[(35/3)]+1 =(70/3)+1 =73/3
Anusha 27 days ago
How is
Hi, This is a splecial case and you can see it is of form f(x) g(x) = 1 where f(x) is 1 + sinx and g(x) = 1/x 2 Direct Solution of above form is e g(x) (f(x) -1) Put f(x) and g(x) and get...
Harsh Patodia one month ago
View all Questions »