
Two A.P’s have the same common difference. The first term of one A.P. is 3, and that of the other is 8. What is the difference between their
(i) 2nd
(ii) 4th
(iii) 10th
(iv) 30th terms respectively.
Answer
615.3k+ views
Hint: The difference between any two corresponding terms of such AP’s is the same as the difference between their first terms.
Complete step-by-step answer:
Let the common difference of the two A.P’s be d.
It is given that the common difference of both A.P is same and
$ \Rightarrow $ Let first A.P be denoted as A.
$ \Rightarrow $ Second A.P be denoted as B.
$ \Rightarrow $ First term of the first A.P is given as ${A_1} = 3$
$ \Rightarrow $ First term of the second A.P is given as ${B_1} = 8$
We know that, kth term of any A.P = first term of that A.P. + (k-1)(common difference of that A.P.)
So, solving all four parts of the question,
i) Let the 2nd term of the first A.P is ${A_2}$.
$ \Rightarrow $${A_2} = {A_1} + (2 - 1)d = 3 + d$
Let the 2nd term of the second A.P is ${B_2}$.
$ \Rightarrow {B_2} = {B_1} + (2 - 1)d = 8 + d$
So, difference between the 2nd term of both A.P’s will be
$\Rightarrow{B_2} - {A_2} = (8 + d)-(3 + d) = 5$
ii) Let the 4th term of the first A.P is ${A_4}$.
$ \Rightarrow {A_4} = {A_1} + (4 - 1)d = 3 + 3d$
Let the 4th term of the second A.P is ${B _4}$.
$ \Rightarrow {B_4} = {B_1} + (4 - 1)d = 8 + 3d$
So, difference between the 4th term of both A.P’s will be
$ \Rightarrow {B_4} - {A_4} = (8 + 3d) - \left( {3 + 3d} \right) = 5$
iii) Let the 10th term of the first A.P is ${A_{10}}$.
\[ \Rightarrow {A_{10}} = {A_1} + (10 - 1)d = 3 + 9d\]
Let the 10th term of the first A.P is ${B_{10}}$.
\[ \Rightarrow {B_{10}} = {B_1} + (10 - 1)d = 8 + 9d\]
So, difference between the 10th term of both A.P’s will be
$ \Rightarrow {B_{10}} - {A_{10}} = (8 + 9d) - (3 + 9d) = 5$
iv) Let the 30th term of the first A.P is ${A_{30}}$.
\[ \Rightarrow {A_{30}} = {A_1} + (30 - 1)d = 3 + 29d\]
Let the 30th term of the first A.P is ${B_{30}}$.
\[ \Rightarrow {B_{30}} = {B_1} + (30 - 1)d = 8 + 29d\]
So, difference between the 30th term of both A.P’s will be
$ \Rightarrow {B_{30}} - {A_{30}} = (8 + 29d) - \left( {3 + 29d} \right) = 5$
Hence the difference between the nth term of these two A.P’s will always be equal to 5.
Note: Whenever you come up with these types of problems then find what is asked for the nth term (like the difference of nth term of both A.P’s in this case) and then directly substitute the value of n. This way the problem can be solved faster. In this case the difference of the nth terms doesn’t depend on n and hence we get the same difference irrespective of the value of n.
Complete step-by-step answer:
Let the common difference of the two A.P’s be d.
It is given that the common difference of both A.P is same and
$ \Rightarrow $ Let first A.P be denoted as A.
$ \Rightarrow $ Second A.P be denoted as B.
$ \Rightarrow $ First term of the first A.P is given as ${A_1} = 3$
$ \Rightarrow $ First term of the second A.P is given as ${B_1} = 8$
We know that, kth term of any A.P = first term of that A.P. + (k-1)(common difference of that A.P.)
So, solving all four parts of the question,
i) Let the 2nd term of the first A.P is ${A_2}$.
$ \Rightarrow $${A_2} = {A_1} + (2 - 1)d = 3 + d$
Let the 2nd term of the second A.P is ${B_2}$.
$ \Rightarrow {B_2} = {B_1} + (2 - 1)d = 8 + d$
So, difference between the 2nd term of both A.P’s will be
$\Rightarrow{B_2} - {A_2} = (8 + d)-(3 + d) = 5$
ii) Let the 4th term of the first A.P is ${A_4}$.
$ \Rightarrow {A_4} = {A_1} + (4 - 1)d = 3 + 3d$
Let the 4th term of the second A.P is ${B _4}$.
$ \Rightarrow {B_4} = {B_1} + (4 - 1)d = 8 + 3d$
So, difference between the 4th term of both A.P’s will be
$ \Rightarrow {B_4} - {A_4} = (8 + 3d) - \left( {3 + 3d} \right) = 5$
iii) Let the 10th term of the first A.P is ${A_{10}}$.
\[ \Rightarrow {A_{10}} = {A_1} + (10 - 1)d = 3 + 9d\]
Let the 10th term of the first A.P is ${B_{10}}$.
\[ \Rightarrow {B_{10}} = {B_1} + (10 - 1)d = 8 + 9d\]
So, difference between the 10th term of both A.P’s will be
$ \Rightarrow {B_{10}} - {A_{10}} = (8 + 9d) - (3 + 9d) = 5$
iv) Let the 30th term of the first A.P is ${A_{30}}$.
\[ \Rightarrow {A_{30}} = {A_1} + (30 - 1)d = 3 + 29d\]
Let the 30th term of the first A.P is ${B_{30}}$.
\[ \Rightarrow {B_{30}} = {B_1} + (30 - 1)d = 8 + 29d\]
So, difference between the 30th term of both A.P’s will be
$ \Rightarrow {B_{30}} - {A_{30}} = (8 + 29d) - \left( {3 + 29d} \right) = 5$
Hence the difference between the nth term of these two A.P’s will always be equal to 5.
Note: Whenever you come up with these types of problems then find what is asked for the nth term (like the difference of nth term of both A.P’s in this case) and then directly substitute the value of n. This way the problem can be solved faster. In this case the difference of the nth terms doesn’t depend on n and hence we get the same difference irrespective of the value of n.
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