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Which term of the A.P. 3,15,27,39,....will be 120 more than its ${21^{st}}$ term?

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Hint: Find out the 21st term of the series by using the nth term formula and add 120 to the calculated 21st term. Proceed to find out n for the calculated nth value.


As we know that the given A.P is 3,15,27,39,.... 

Here, Common difference, d = difference of any two consecutive terms = 15 - 3 = 12. 

As we know that the ${n^{th}}$ term of the A.P is given by ${t_n}$

   $\Rightarrow {\text{So, }}{t_n} = a + (n - 1) \times d{\text{ ,}}\;{\text{where}}\;a = 3,d = 12{\text{ (Eq 1)}} \\$


Hence first we have to find the value of ${21^{st}}$ term of the given A.P

$ \therefore {\text{ }}{t_{21}} = 3 + (21 - 1) \times 12$

$= 3 + 20 \times 12$

$= 3 + 240$

$\Rightarrow \therefore {\text{ }}{t_{21}} = 243$


Let the term will be ${m^{th}}$ which is 120 greater than its ${21^{st}}$  term.

So, according to equation 1,

So, ${t_m} = 3 + (m - 1) \times 12$

$\Rightarrow {\text{ }}{t_m} = 12m - 9\;{\text{ (Eq 2)}}$


And it is given that ${t_m} = 120 + {t_{21}}{\text{ (Eq 3)}}$

So, on comparing equation 2 and 3 and putting the value of ${t_{21}}$ in it we get,

   $\Rightarrow 12m - 9 = 120 + 243 = 363$

   $\Rightarrow m = \dfrac{{372}}{{12}} = 31$


Hence, ${31^{st}}$ term of the given A.P is 120 more than its ${21^{st}}$ term.


NOTE: - Whenever you come up with this type of problem the best way is to compute the given term of that A.P whose relation is given in the question and then compare that with the required condition.