Sum

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

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#### Solution

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, a_{n} = a + (n -1)d

So, for 21^{st} term (n = 21),

a_{21} = 3 + (21 - 1)(12)

= 3 +20(12)

= 3 + 240

= 243

Let us take the term which is 120 more than the 21^{st} term as a_{n}.

a_{n }= 120 + a_{21}

So, a_{n }= 120 + 243

a_{n }= 363

Also, a_{n }= a +(n - 1)d

363 = 3 + (n -1)12

363 = 3 + 12n - 12

363 = -9 = 12n

363+ 9 = 12n

Further simplifying, we get

372 = 12n

`"n" = (370)/(12)`

n = 31

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

Concept: Algebraic Conditions for Number of Solutions

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