IIT JEE , AIEEE Forum - Mathematics , Algebra

deedee

a,b.l.....r respectively d 1st ...2nd....& last termz of a gp............

d sum is..............

aadha answer aa raha hai........aadha ni aaa raha

netkid07

first term=a

second term=b=ar where r is common difference......

r=b/a

last term= l =ar^(n-1) where n is total no. of terms................

r^n=lr/a=lb/a^2

put in the formula for sum of a g.p

=a(1-r^n/1-r)

solve and u get:::::

(a^2-lb/a-b)

rate if u like..............

madman

b=aR
R=b/a
l=aR^(n-1)
lR=aR^n
R^n=lR/a

sum=a(R^n-1)/(r-1)
=a(lR/a-1)/(r-1)
=a(((lb-a2)/a2)/(b-a)/a)
=a((lb-a2/a)/(b-a))
=lb-a^2/(b-a)
same as netkid,s

Greatdreams

ya i solved it ans is (a^2-lb/a-b).There is no diff btw madman's and my ans you multiply -1 to numeratr and deno.You will get the ans

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