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# ${\text{Which term of the }}A.P.\;3,15,27,39,....{\text{will be }}120{\text{ }}{\text{more than its }}{21^{st}}{\text{term}}?$ Verified
${\text{As we know that the given A}}{\text{.P is }} \\ \Rightarrow A.P:\;3,15,27,39,.... \\ {\text{Common difference}},\;d = {\text{difference of any two consecutive terms}} = 15 - 3 = 12. \\ {\text{As we know that the }}{n^{th}}{\text{ term of the A}}{\text{.P is given by }}{t_n}. \\ \Rightarrow {\text{So, }}{t_n} = a + (n - 1) \times d{\text{ ,}}\;{\text{where}}\;a = 3,d = 12{\text{ (1)}} \\ {\text{Hence first we have to find the value of }}{21^{st}}{\text{ term of the given A}}{\text{.P}}{\text{.}} \\ \therefore {\text{ }}{t_{21}} = 3 + (21 - 1) \times 12\; \\ {\text{ }} = 3 + 20 \times 12\;{\text{ }}\; \\ {\text{ }} = 3 + 240 \\ \Rightarrow \therefore {\text{ }}{t_{21}} = 243 \\ {\text{Let the term will be }}{m^{th}}{\text{ which is }}120{\text{ greater than its }}{21^{st}}{\text{ term }} \\ {\text{So, according to equation 1}} \\ {\text{So, }}{t_m} = 3 + (m - 1) \times 12\; \\ \Rightarrow {\text{ }}{t_m} = 12m - 9\;{\text{ (2)}}\; \\ {\text{And it is given that }}{t_m} = 120 + {t_{21}}{\text{ (3)}} \\ {\text{So, on comparing equation }}2{\text{ }}and{\text{ }}3{\text{ and putting the value of }}{t_{21}}{\text{ in it }}{\text{we get}}, \\ \Rightarrow 12m - 9 = 120 + 243 = 363 \\ \Rightarrow m = \dfrac{{372}}{{12}} = 31 \\ {\text{Hence, }}{31^{st}}{\text{ term of the given A}}{\text{.P is 120 more than its 2}}{1^{st}} \\ {\text{NOTE: - Whenever you came up with this type of problem them best way is to to compute}} \\ {\text{the given term of that A}}{\text{.P whose relation is given in the question and then compare that with}} \\ {\text{required condition}}{\text{.}} \\$