The difference between $a_{30}$ and $a_{20}$ for the AP $- 3, - 7, - 11$ is $^{'} d^{'}$ , then $\frac{| d |}{10}$ equals

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Solution

We know that the $n$ th term of A.P is $a_{n} = a + (n - 1) d$ where $a$ is the first term, $n$ is the number of terms and $d$ is the common difference.

In the given A.P, the first term is $a = - 3$ and the common difference is:

$d = a_{2} - a_{1} = - 7 - (- 3) = - 7 + 3 = - 4$

Now, we find $a_{20}$ and $a_{30}$ as follows:

$a_{20} = - 3 + (20 - 1) (- 4) = - 3 + (19 \times - 4) = - 3 - 76 = - 79$

$a_{30} = - 3 + (30 - 1) (- 4) = - 3 + (29 \times - 4) = - 3 - 116 = - 119$

Subtract $a_{20}$ from $a_{30}$ as shown below:

$a_{30} - a_{20} = - 119 - (- 79) = - 40$

So, the difference between $a_{20}$ and $a_{30}$ is $- 40$ .
$∴$ $\frac{| d |}{10}$ = $4$ .

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For the AP –3, –7, –11,… can we find directly $a_{30} - a_{20}$ without actually finding $a_{30} a n d a_{20}$ ? Give the reason for your answer.

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