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प्रश्न
Which term of the A.P. 8, 14, 20, 26, ... will be 72 more than its 41st term?
उत्तर
In the given problem, let us first find the 41st term of the given A.P.
A.P. is 8, 14, 20, 26 …
Here,
First term (a) = 8
Common difference of the A.P. (d) = 14 - 8 = 6
Now as we know
`a_n = a+(n + 1)d`
So for 41st term (n = 41)
`a_(41) = 8 + (41 - 1)(6)`
= 8 + 40(6)
= 8 + 240
= 248
Let us take the ter which is 72 more than the 41st term as an So
`a_n = 72 + a_41`
= 72 + 248
= 320
Also `a_n = a + (n - 1)d`
320 = 8 + (n - 1)6
320 = 8 + 6n - 6
320 = 2 + 6n
320 - 2 = 6n
Futher simplifying weget
318 = 6n
`n = 318/6`
n = 53
Therefore the 53rd term of the given A.P is 72 more than the the 41st term.
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