For what value of n are the nth term of the following two ap are same 13 19,25 and 69,68,67

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