Question #04343

2 Answers
Dec 9, 2017

#333#

Explanation:

First of all, we have that the sum of all the odd integers from #28# to #46# are #29 + 31 + ... + 43 + 45#. That is a total of #9# numbers.

Let's represent this sum as a variable #s#:
#s = 29 + 31 + ... + 43 + 45#

Then reverse the order:
#s = 45 + 43 + ... + 31 + 29#

Add these two each other:
#s + s = (29 + 45) + (31 + 43) + ... + (43 + 31) + (45 + 29)#

#s + s# becomes #2s#. Notice, from the left to right of the first two terms. As #29# increases by #2# to become #31#, we can see that #45# also decreases by #2# to become #43#. Since this goes on all the way, this means that essentisally, all of these terms are the same:
#29 + 45 = 31 + 43 = ... = 43 + 31 = 45 + 29#

And since there are #9# of them, we could just sum one of these terms and multiply by #9#:
#2s = (29 + 45) * 9 = 74 * 9 = 666#.

Finally, we divide both sides by #2#:
#{2s}/2 = 666/2#

#s = 333#

Dec 9, 2017

#333#

Explanation:

using Arithmatic progression
#a_1# #=29#
#a_n# #=45#
we know that
#a_n = a_1 + (n-1)d#
#45=29+(n-1)d#
#45-29=(n-1)d#
#16=(n-1)d# {here d=2 because we have to count only odd number }
#16/2=n-1#
#8=n-1#
#8+1=n#

now, #S_n=n/2 (a_1 +a_n)#
#S_9=9/2 (29+45)#
#S_9=9/2 (74)#
#S_9=9×37 = 333#
hope ti helps