CAT 2019

Slot 2

QA

Question & Solution

$a_1 - a_2 + a_3 - a_4 + .... + (-1)^{n - 1} a_n = n$

It is clear from the above equation that when n is odd, the co-efficient of a is positive otherwise negative.

$a_1 - a_2 = 2$

$a_1 = a_2 + 2$   

$a_1 - a_2 + a_3 = 3$

On substituting the value of $a_1$ in the above equation, we get

$a_3$ = 1

$a_1 - a_2 + a_3 - a_4 = 4$

On substituting the values of $a_1, a_3$ in the above equation, we get

$a_4$ = -1

$a_1 - a_2 + a_3 - a_4 +a_5 = 5$

On substituting the values of $a_1, a_3, a_4$ in the above equation, we get

$a_5$ = 1

So we can conclude that $a_3, a_5, a_7....a_{n+1}$ = 1 and $a_2, a_4, a_6....a_{2n}$ = -1

Now we have to find the value of $a_{51} + a_{52} + .... + a_{1023}$

Number of terms = 1023=51+(n-1)1

n=973

There will be 486 even and 487 odd terms, so the value of $a_{51} + a_{52} + .... + a_{1023}$ = 486*-1+487*1=1