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

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

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