GMAT CLUB OLYMPICS: If a + a/(1 + 2)

Hello everyone

Here is my solution.
The correct answer is 51 which is B

Here the Nth term is = a/(1+2+3+..+n)

Sum of first n natural nos. = n(n+1)/2
Therefore the nth term becomes = 2a/(n(n+1))
=2a[1/n - 1/(n+1)]
On summation of these terms till 101, all the middle terms get cancelled and we are left with 2a[1/1 - 1/101]
Now, 2a[1/1 - 1/101] = 101
=> 2a*101/102 = 101
=> a = 51

Hence Option B is the correct answer.

We have:
1+2+...+n = (n+1)n/2
=> 1/(1+2+...+n) = 2/(n+1)n = 2*(1/n - 1/(n+1))

Thus we have:
a/1 + a/(1+2) + ... + a/(1+2+...+101) = 101
=> a*(1/1 + 1/(1+2) + ... + 1/(1+2+...+101)) = 101
=> 2a*(1/1 - 1/2 + 1/2 - 1/3 +... + 1/101 - 1/102) = 101
=> 2a*(1 - 1/102) = 101
=>2a*101/102 = 101
=> 2a = 102
=> a = 51

The answer is B

a(1 + 1/3 +1/6 +1/10 + ...1/(1+...101) = 101
because the sum (1/3 +1/6 +1/10 + ...1/(1+...10)) < 1 => 1 + sum < 2 => a(1 + sum) < 2a
<=> 101 < 2a
<=> 101/2 < a
<=> 50.5 < a
=> a = 51

The moment you see the formula of nth number, it should tell you we are looking at a sequence ..
Also, the denominator seems to be increasing in a certain pattern. So, let us try to find out.

Here the nth term is \(\frac{a}{\frac{n(n+1)}{2}}=\frac{2a}{n(n+1)}=2a(\frac{1}{n}-\frac{1}{n+1})\)

\(a + \frac{a}{(1 + 2)} + \frac{a}{(1 + 2 + 3)} + ... + \frac{a}{(1 + 2 + 3 + ... + 101)} = 101\)

\(2a(\frac{1}{1*2}+\frac{1}{(1 + 2)*2} + \frac{1}{(1 + 2 + 3)*2} + ... + \frac{1}{(1 + 2 + 3 + ... + 101)*2} = 101\)

\( \frac{1}{(1 )*2} =\frac{1}{1*2}= \frac{1}{1}-\frac{1}{2}\)..

\( \frac{1}{(1 + 2)*2} =\frac{1}{3*2}= \frac{1}{2}-\frac{1}{3}\).

\( \frac{1}{(1 + 2+3)*2} = \frac{1}{3}-\frac{1}{4}\)..
.....
\( \frac{1}{(1 +2+..+101)*2} =\frac{1}{101*102}= \frac{1}{101}- \frac{1}{102}\).

ADD all
\(2a(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}..... \frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}) = 2a(1-\frac{1}{102}) =2a( \frac{101}{102})=101\)
\(a=51\)


B

You can solve it by using the knowledge of just this one formula - Sum of n consecutive terms from 1 to n is n(n+1)/2

Step 1 - Take the common term a out of the expression : a * (1 + \(\frac{1}{1+2} \) + \(\frac{1}{1+2+3} \) + ... + \(\frac{1}{1+2+3+..101} \) ) = 101

Step 2 - Substitute the terms with the formula : a x (1 + \(\frac{2}{2 * 3} \) + \(\frac{2}{3 * 4} \) + ... + \(\frac{2}{101 * 102} \) ) = 101

Step 3 - Write 1 as 2/2 to create standard form : a * (\(\frac{2}{1 * 2} \) + \(\frac{2}{2 * 3} \) + \(\frac{2}{3 * 4} \) + ... + \(\frac{2}{101 * 102} \) ) = 101

Step 4 - Take common term 2 out : a * 2 * (\(\frac{1}{1 * 2} \) + \(\frac{1}{2 * 3} \) + \(\frac{1}{3 * 4} \) + ... + \(\frac{1}{101 * 102} \) ) = 101

Now, \(\frac{1}{n * (n+1)} \) can be written as \(\frac{(n+1) - n }{n * (n+1)} \) which can further be broken down into ( \(\frac{1}{n} \) - \(\frac{1}{(n+1)} \) )

Step 5 - Substitute the expression with the from above : a * 2 * (\(\frac{1}{1} \) - \(\frac{1}{2} \) + \(\frac{1}{2} \) - \(\frac{1}{3} \) + \(\frac{1}{3} \) + ... - \(\frac{1}{101} \) + \(\frac{1}{101} \) - \(\frac{1}{102} \)) = 101

We can observe that all terms in the fractional expression cancel out except the first and the last terms.

Step 6 - Cancel out terms in the fractional expression : a * 2 * (\(\frac{1}{1} \) - \(\frac{1}{102} \)) = 101

Step 7 - Find out a by cancelling out common term 101 on LHS and RHS : a * 2 * (\(\frac{101}{102} \) ) = 101

( a *2 * \(\frac{1}{102} \) )= 1

Answer -> a=51

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