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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. |
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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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