Top 125 GMAT Quant Questions PDF 2025, Important Formulas

$$f(x)=ax^2+bx+c$$, such that 1 and -1 are the roots of f(x)

This means that $$f(1)=0$$ and $$f(-1)=0$$

$$f(1)=a(1)^2+b(1)+c=0$$ -> $$a+2b+c=0\rightarrow1$$

$$f(-1)=a(-1)^2+b(-1)+c=0$$ -> $$a-2b+c=0\rightarrow2$$

Adding eq. 1 and eq. 2, we will get $$a+c=0$$ or $$a=-c$$

Subtracting eq. 2 from eq. 1, we will get $$4b=0$$ or $$b=0$$.

Therefore, the equation becomes -

$$f(x)=ax^2-a=a(x^2-1)$$ or $$f(x)=-cx^2+c=-c(x^2-1)$$

f(2) = 3a or f(2) = -3c

f(5) = 24a or f(5) = -24c. This is a multiple of 6.

f(4) = 15a or f(4) = - 15c

f(3) = 8a or f(3) = -8c

Thus, option E can never be TRUE.

$$T_n=\ \dfrac{\ 1}{n\left(n+1\right)}$$

$$T_n=\dfrac{\ 1}{n}-\dfrac{\ 1}{n+1}$$

$$S_{20\ }=\ T_{1\ }+T_2+T_3+....T_{20}$$
=  $$1-\dfrac{\ 1}{2}+\dfrac{\ 1}{2}-\dfrac{\ 1}{3}+\ \dfrac{\ 1}{3}-\dfrac{\ 1}{4}+.......+\ \dfrac{\ 1}{20}-\dfrac{\ 1}{21}$$

$$S_{20}\ =1-\dfrac{\ 1}{21}=\dfrac{\ 20}{21}$$

The amount saved by Ousman in the first week is given to be $5, since we are given that every week he saves $2 more than the previous week, the amount saved by him every week would be as follows,

$5, $7, $9,... for 52 weeks.

We can clearly see that the terms are in AP with a common difference of 2, and there are 52 terms in total in the sequence. The sum of these terms can be calculated as,

$$S_n\ =\ \dfrac{n}{2}\left(2a\ +\ \left(n\ -\ 1\right)d\right)$$

$$S_{52}\ =\ \frac{52}{2}\left(2\left(5\right)\ +\ \left(52\ -\ 1\right)2\right)\ =\ 26\left(10\ +\ 102\right)\ =\ 2912\ $$

So, the total amount saved throughout the year is $2912.

Hence, the correct answer is option B.

We are given $$a_1=3$$

$$a_2=2a_1-5\ =\ 6-5\ =\ 1$$

$$a_3=2a_2-5\ =\ 2-5\ =\ -3$$

$$a_4=2a_3-5\ =\ -6-5\ =\ -11$$

$$a_5=2a_4-5\ =\ -22-5\ =\ -27$$

$$a_6=2a_5-5\ =\ -54-5\ =\ -59$$

We need to find $$a_6-a_5=-59-\left(-27\right)\ =\ -59+27\ =-32$$

The sum of n terms of a series is given as $$S_n=2n^2+5n$$.

We know that $$T_n=S_n-S_{n-1}=2n^2+5n-\left(2\left(n-1\right)^2+5\left(n-1\right)\right)$$

$$T_n=2n^2+5n-\left(2n^2+n-3\right)$$

$$T_n=4n+3$$

Therefore, the 14th and 6th terms will be equal to

$$T_{14}=4\left(14\right)+3\ =\ 59$$

$$T_6=4\left(6\right)+3\ =\ 27$$

Difference = 59-27 = 32