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Question

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

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Solution

Given sequence is 3,15,27,39,...

The first term is a=3

The common difference is d=153=12

we know that the nth term of the arithmetic progression is given by a+(n1)d

Let the mth term be 120 more than the 21st term

Therefore, 120+mthterm=21stterm

120+a+(m1)d=a+(211)d

120+(m1)12=(20)12

12m12=240+120

12m=360+12

12m=372

m=37212=31

Therefore, the 31st term is 120 more than the 21st term

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