How do you find the sum of #6-3/4i# from i=1 to 22?

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Answer 1 · 2017-07-29T06:21:34

Sum is #-57 3/4#

As the #i^(th)# term is #6-3/4i#,

first term is #6-3/4=5 1/4=21/4#

next few terms are #18/4,15/4,.......#

Hence, it an arithmetic sequence with first term as #a_1=21/4# and common difference as #-3/4#

Now sum of an arithmetic sequence up tp #n# terms, whose first term is#a# and common difference is #d# is #S_n=n/2(2a+d(n-1))#.

Hence sum of first #22# terms is

#S_22=22/2(2xx21/4+(-3/4)xx21)#

#=11(42/4-63/4)#

#=11xx(-21/4)=-231/4=-57 3/4#