For the A.P -3, -7, -11,..... can we directly find a_{30} - a_{20} without actually finding a_{30} and a_{20}? Give reasons for your answer.

Yes, we have common difference as -7 - (-3) = -4

a_{30} - a_{29} = d -- > (a_{n} - a_{(n - 1)} = d) (the difference b/w 2 consecutive terms of an AP is the common difference)

a_{29 }= a_{30} - d --- (1)

Now let's take

a_{29} - a_{28} = d --- (2)

Substituting (1) in (2)

a_{30} - d - a_{28} = d

a_{30} - a_{28} = 2d

From these cases, we can arrive to the conclusion that

a_{n} - a_{(n - x)} = x * d

Where a_{n} is the n^{th} term,

x is the no. of terms subtracted

and d is the common difference

a_{30} - a_{20} = a_{30} - a_{(30 - 10)}

That is of the form a_{n} - a_{(n - x)}

Therefore

a_{30} - a_{(30 - 10)} = 10 d

10d = 10 * -4 = -40

Therefore the answer is -40.

We did this without calculating the individual values for a_{30} and a_{20}

(P.S = I want a MeritNation Expert to tell me if such a formula exists.... I've heard of a_{n} - a_{(n - 1)} = d, I improved on it)

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