For the A.P -3, -7, -11,..... can we directly find a30 - a20 without actually finding a30 and a20? Give reasons for your answer.
Yes, we have common difference as -7 - (-3) = -4
a30 - a29 = d -- > (an - a(n - 1) = d) (the difference b/w 2 consecutive terms of an AP is the common difference)
a29 = a30 - d --- (1)
Now let's take
a29 - a28 = d --- (2)
Substituting (1) in (2)
a30 - d - a28 = d
a30 - a28 = 2d
From these cases, we can arrive to the conclusion that
an - a(n - x) = x * d
Where an is the nth term,
x is the no. of terms subtracted
and d is the common difference
a30 - a20 = a30 - a(30 - 10)
That is of the form an - a(n - x)
Therefore
a30 - a(30 - 10) = 10 d
10d = 10 * -4 = -40
Therefore the answer is -40.
We did this without calculating the individual values for a30 and a20
(P.S = I want a MeritNation Expert to tell me if such a formula exists.... I've heard of an - a(n - 1) = d, I improved on it)