Question

Which term of the AP 3 , 15 , 27 , 39 will be 120 more than its 21st term ? ​

Answer 1

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

The first term is a = 3

The common difference is d = 15 − 3 = 12

we know that the nth term of the arithmetic progression is given by a+(n−1)d

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

Therefore, 120 + mth term = 21st term

⟹ 120 + a + (m − 1)d = a + (21 − 1)d

⟹ 120 + (m − 1)12 = (20)12

⟹ 12m − 12 = 240 + 120

⟹ 12m = 360 + 12

⟹ 12m = 372

⟹ m = 372/12 = 31

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

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Answer 2

In the given problem, let us first find the 21st term of the given A.P.

A.P. is 3, 15, 27, 39 …

Here

First term (a) = 3

Common difference of the A.P. (d) = 15 - 3 = 12

Now, as we know,

an=a+(n-1)d

So for 21st term (n = 21)

a21=3+(21-1)(12)

= 3 + 20(12)

= 3 + 240

= 243

Let us take the term which is 120 more than the 21st term as an

So,

an=120+a21

a21=3+(21-1)(12)

=3+20(12)

= 3 + 240

= 243

Let us take the term which is 120 more than the 21st term as an so

an=120+a21

= 120 + 243

= 363

Also an = a ( n-1 ) d

an=a+(n-1)d

363 = 3 + (n - 1)12

363 = 3 + 12n - 12

363 + 9 = 12n

Further simplifying we get

372 = 12n

n=370/12

n = 31

Therefore the 31st term of the given A.P is 120 more than the 21st term.

Hope it helps uh ❤️༄