Two A.P.'s have the same common difference.

Question:

Two A.P.'s have the same common difference. The first term of one of these is 8 and that of the other is 3. The difference between their 30th term is

(a) 11

(b) 3

(c) 8

(d) 5

Solution:

Here, we are given two A.P.’s with same common difference. Let us take the common difference as d.

Given,

First term of first A.P. (a) = 8

First term of second A.P. (a) = 3

We need to find the difference between their 30th terms.

So, let us first find the 30th term of first A.P.

$a_{30}=a+(30-1) d$

$=3+29 d$ $\ldots(1)$

Similarly, we find the 30th term of second A.P.

$a_{30}^{\prime}=a^{\prime}+(30-1) d$

$=3+29 d$ ........(2)

Now, the difference between the $30^{\text {th }}$ terms is,

$a_{30}-a_{30}^{\prime}=(8+29 d)-(3+29 d)$

$=8+29 d-3-29 d$

$=8-3$

$=5$

Therefore, $a_{30}-a_{30}^{\prime}=5$

Hence, the correct option is (d).

 

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