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For what value of n are the nth terms of the following two Aps the same 13, 19, 25, … and 69, 68, 67,…? Also, find this term.

Answer» Let nth terms of the given progressions be `t_(n) "and" T_(n)` respectively.
The first AP is 13, 19, 25, ….
Let its first term be a and common difference be d. Then,
a = 13 and d = (19-13) = 6.
So, its nth term is given by
`t_(n) = a + (n-1)d`
`rArr t_(n) = 13 + (n-1) xx 6`
`rArr t_(n) = 6n + 7 " "...(i)`
The second AP is 69, 68, 67, ...
Let its first term be A and common difference be D. Then,
A = 69 and D = (68-69) =-1.
So, it nth term is given by
`T_(n) = A + (n-1) xx D`
`rArr T_(n) = 69 + (n-1) xx (-1)`
`rArr T_(n) = 70-n`
Now, `t_(n) = T_(n) rArr 6n + 7 = 70-n`
`rArr 7n = 63 rArr n = 9`
Hence, the 9th term of each AP is the same
This term ` = 70-9 = 61 [ because T_(n) = (70 -n)].`


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