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4 Answers
Dec 9, 2017

D #100162#

Explanation:

Given:

#46+49+52+...+775#

Note that the common difference is #3#, so the number of terms is:

#(775-46)/3 + 1 = 729/3 + 1 = 243+1 = 244#

Since this is an arithmetic series, the average term is the same as the average of the first and last terms, namely:

#(46+775)/2 = 821/2#

So the sum of the series is:

#244 * 821/2 = 122 * 821 = 100162#

Dec 9, 2017

See the answer below...

Explanation:

For any AP Series, if #"first term"# #-># a
#" common difference "->d#,

  • The #color(red)(n_(th)# term of the series is #color(red)(n_(th) "term"=a+(n-1)d#
  • Sum of the series upto #color(red)(n_(th)# term is #color(red)("sum"=n/2{2a+(n-1)d}#

In the given series, #color(red)(a=46# and #color(red)(d=(49-46)=3#
Hence, #" "color(red)(46+(n-1)3=775#
#" "=>3n-3=729=>n=244#

Now, we have the value of #a,d,n# ,so we can easily determine the result of #color(red)("sum"#.

#color(red)("sum")=244/2{2xx46+(244-1)3}=color(green)(100162#

Hence,the correct option id (D)#100162#

Hope it helps...
Thank you...

Dec 9, 2017

#D#

Explanation:

#"the sum to n terms of an arithmetic series is"#

#•color(white)(x)S_n=n/2[2a+(n-1)d]#

#"where a is the first term and d "color(blue)"the common difference"#

#"here "a=46" and "d=52-49=49-46=3#

#"the nth term of an arithmetic series is"#

#•color(white)(x)a_n=a+(n-1)d#

#rArr46+3(n-1)=775larrcolor(blue)"solve for n"#

#rArr46+3n-3=775#

#rArr3n=732rArra=732/3=244#

#"we require the sum to 244 terms"#

#rArrS_(244)=122[(2xx46)+(243xx3)]#

#color(white)(rArrS_(244))=100162#

Dec 9, 2017

Figure out which term #775# is, add the reverse of the sum to itself, transform it into multiplication, and obtain D. #100162#

Explanation:

Let's set that sum into a variable, say, #M#:

#M = 46 + 49 + 52 + ... + 775#

OK, now let's use the formula for arithmetic sequences to find out which term #775# is, or, well, how many terms are in this sum?

#a_n = a_1 + d(n - 1)#

We know #a_n = 775# since that's the number whose #n# we're currently looking for... we know the first term, #a_1 = 46#... then there's the difference between two terms, say between #a_2# and #a_1#: #d = a_2 - a_1 = 49 - 46 = 3#. Substitute:

#775 = 46 + 3(n - 1)#

Expand #3(n - 1)#:

#775 = 46 + 3n - 3#

Add the constants:

#775 = 43 + 3n#

Subtract both sides by #43#:

#775 - 43 = 43 + 3n - 43#

#732 = 3n#

Divide both sides by #3#:

#732/3 = (3n)/3#

#244 = n#

So now we know that #775# is the #244#th term of the arithmetic sequence. Bigger than expected! Now back to #M#:

#M = 46 + 49 + 52 + ... + 775#

Let's reverse the sum:

#M = 775 + ... + 52 + 49 + 46#

Add those two together:

#M + M = (46 + 775) + ... + (775 + 46)#

The left side simplifies to #2M#, the right side... well, let's reveal a bit more about the sum (by remembering about the common difference, #d = 3#):

#2M = (46 + 775) + (49 + 772) + ... + (772 + 49) + (775 + 46)#

Seems symmetrical! Not only that, but all these terms are equal; as #46# is added by #3# to become #49#, #775# is also decreased by #3# to become #772#. This keeps going until it becomes #(775 + 46)# at the end.

And how many terms are there? #244#.

#2M = (46 + 775)(244)#

See what I did there? Since they're all equal, I could just use one of them, and multiply it over by the amount of terms. However, we are currently solving for #2M#, not #M#, so let's divide by #2# first:

#(2M)/2 = ((46 + 775)(244))/2#

#M = (46 + 775)(122)#

Solve!

#M = (821)(122)#

Ouch. In this case I would usually "split" one of the numbers, usually the smaller one, then expand, as such:

#M = (821)(100 + 20 + 2)#

#M = (821)(100) + (821)(20) + (821)(2)#

#M = 82100 + 16420 + 1642#

#M = 98520 + 1642#

#M = 100162#

Done!