For the $AP: -3, -7, -11, \ldots,$ can we find $a_{30} - a_{20}$ directly without actually finding $a_{30}$ and $a_{20}$? Give reasons for your answer.
(A) Yes,it is possible to find the value directly.
The $n$-th term of an $AP$ is given by $a_n = a + (n - 1)d$.
Therefore,$a_{30} = a + (30 - 1)d = a + 29d$ and $a_{20} = a + (20 - 1)d = a + 19d$.
Subtracting the two terms: $a_{30} - a_{20} = (a + 29d) - (a + 19d) = 10d$.
From the given $AP$,the common difference $d = -7 - (-3) = -7 + 3 = -4$.
Substituting the value of $d$ into the expression: $a_{30} - a_{20} = 10(-4) = -40$.
The $n$-th term of an $AP$ is given by $a_n = a + (n - 1)d$.
Therefore,$a_{30} = a + (30 - 1)d = a + 29d$ and $a_{20} = a + (20 - 1)d = a + 19d$.
Subtracting the two terms: $a_{30} - a_{20} = (a + 29d) - (a + 19d) = 10d$.
From the given $AP$,the common difference $d = -7 - (-3) = -7 + 3 = -4$.
Substituting the value of $d$ into the expression: $a_{30} - a_{20} = 10(-4) = -40$.
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