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प्रश्न
Which term of the A.P. 3, 15, 27, 39, ... will be 120 more than its 21st term?
उत्तर
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 21st 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 21st term as `a_n` So,
`a_n = 120 + a_21`
`a_21 = 3 + (21 - 1)(12)`
`= 3+ 20(12)`
= 3 + 240
= 243
Let us take the term which is 120 more than the 21st term as `a_n` so
`a_n = 120 + a_21`
= 120 + 243
= 363
Also `a_n = a + (n - 1)d`
363 = 3 + (n - 1)12
363 = 3 + 12n - 12
363 + 9 = 12n
Further simplifying we get
372 = 12n
`n = 372/12`
n = 31
Therefore the 31st term of the given A.P is 120 more than the 21st term.
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