The question: In an A.P, the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is -112.
I was able to solve the actual problem easily. But the strange thing is that the common difference came out to be $-6$. That doesn't make any sense. Why would the sum of the 6-10th terms be higher than the sum of the 1-5th terms if the common difference is negative?
On
Designate the common difference as $a$.
The first five terms are $2,2+a,2+2a,2+3a,2+4a$
and their sum is $10+10a$.
The second five terms are $2+5a,2+6a,2+7a,2+8a,2+9a$
and their sum is $10+35a$.
$10+10a=\frac{10+35a}4$
$4(10+10a)=40+40a=10+35a$
$30+5a=0$
$a=-6$
The twentieth term is $2+19\times(-6)=-112$
The greater the magnitude of a negative number, the farther LEFT on the number line it will be. You probably associate "more-leftward-on-the-number-line" with "less." That association only holds true with positive numbers.
We're more used to positive quantities than to negative ones, so when hearing or reading that $a$ is one fourth of $b$, the immediate idea is that $a$ is less than $b$. At least until we've trained ourselves to get rid of the positive bias. What is always true is that if $b\neq 0$, then one fourth of $b$ is closer to $0$ than $b$.
If $s$ is negative, then $4s < s$, and the constraint that the sum of the first five terms of the progression shall be one fourth of the sum of the next five terms forces the common difference of the progression to have the same sign as the sum of the first five terms - if the common difference is positive, the sum of the first five terms is smaller than the sum of the next five, and if the difference is negative, it is greater.