If #(3+5+7+......+n terms)/(5+8+11+.......+10 terms)# = 7, then the value of n is?

1 Answer
Nov 2, 2017

See below.

Explanation:

Something is lacking...

If #(3+5+7+cdots+(n " terms"))/(5+8+11+cdots+(10n " terms")) = 7#, then the value of #n# is?

#N/D = 7# where

#N = sum_(k=1)^n (2k+1) = n + 2 (n(n+1))/2#
#D = sum_(k=1)^(10n)(2+3k)=2*10 n+3 (10n(10n+1))/2# and then

#n + 2 (n(n+1))/2=7(2*10n+3 (10n(10n+1))/2)#

and after simplification

#n(243-1049n)=0# with solutions

#n = 0# or #n=-243/1049#

Now if instead the formulation were

If #(3+5+7+cdots+(n " terms"))/(5+8+11+cdots+(10 " terms")) = 7#, then the value of #n# is?

then the result could be found by solving

#n + 2 (n(n+1))/2=7(2*10+3 (10(10+1))/2)# giving

#n = {-37, 35}#