The 7th term of an Ap is -1 and the 16th term is 17.FIND THE Nth TERM

1.	The 7th term of an Ap is -1 and the 16th term is 17.FIND THE Nth TERM
Answer» AnsWer : - 13 , - 11 , - 9..... SolutioN : LET, FIRST term be a Common difference be d an = a + ( N - 1 ) d. Case 1 The 7th term of an Ap is - 1. $\tt \dagger \: \: \: \: \: a_7 = - 1.$ Case 2. $\tt \dagger \: \: \: \: \: a_{16} = 17$ Let's Find the nth terms. $\tt \dagger \: \: \: \: \: \boxed{\tt a_n = a + (n - 1)d}$ Taking Case 1. $\tt : \implies \tt a_7 = - 1.$ $\tt : \implies a + (7 - 1)d = - 1.$ $\tt : \implies a + 6d = - 1. \: \: \: \: \: - (1)$ Taking case 2. $\tt : \implies a +(16 - 1)d = 17.$ $\tt : \implies a + 15d = 17. \: \: \: \: \: - (2)$ Subtraction, Equation ( 1 ) and ( 2 ) $\tt a + 15d = 17.$ $\tt a + 6d = - 1.$ ________________________________ $\tt : \implies9d = 18.$ $\tt : \implies d = \dfrac{18}{9}$ $\tt : \implies d = 2.$ Now, PUTTING the value of d = 2. ( in Case 1 ) $\tt : \implies a + 6d = - 1.$ $\tt : \implies a + 6(2)= - 1.$ $\tt : \implies a +12 = - 1.$ $\tt : \implies a = - 1 - 12.$ $\tt : \implies a = - 13.$ $\rule{200}3$ Let find nth terms. → an = a + ( n - 1 )d. n = 1 , 2 , 3... a = - 13. d = 2. $\tt \dagger \: \: \: \: \: a_1 = a + (n - 1)d$ $\tt : \implies a_1 = - 13 + (1- 1)2$ $\tt : \implies a_1 = - 13 +0$ $\tt : \implies a_1 = - 13.$ $\rule{200}3$ $\tt : \implies a_2 = - 13 +(2 - 1)2$ $\tt : \implies a_2 = - 13 +2$ $\tt : \implies a_2 = - 11.$ $\rule{200}3$ $\tt : \implies a_3 = - 13 +(3- 1)2$ $\tt : \implies a_3= - 13 +4$ $\tt : \implies a_3 = -9.$ Therefore, the value of nth of AP , - 13 , - 11 , - 9 .... so on.

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