If the 3rd and the 9th terms of an A.P. are 4 and − 8 respectively,

1.	If the 3rd and the 9th terms of an A.P. are 4 and − 8 respectively, then which term of this A.P is zero.
Answer» GIVEN :- 3rd term of A.P is 4. 9th term of A.P is -8. $\\$ TO FIND :- The term when the A.P is 0. $\\$ FORMULA USED :- $\\ \bigstar\boxed{\sf \: a_{n} = a + (n - 1)d } \\$ Here , a(n) → 'n'th term a → First term n → TOTAL number of terms d → Common difference $\\$ SOLUTION :- $\\$ ♦ 3rd term of A.P is 4. We know , a(n) = a + (n-1)d Here , a(n) = 4 n = 3 Putting VALUES we get.. $\implies\sf4 = a + (3 - 1)d \\$ $\\ \implies \sf 4 = a + 2d \: \: \: \: \: \: \: \: \: \: \: \: - - - (i)\\ \\$ ♦ 9th term of A.P is -8. Now with the same formula ...we get , $\\ \implies \sf - 8 = a + (9 - 1)d \\$ $\\ \implies \sf \: - 8 = a + 8d \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - - - (<klux>II</klux>) \\ \\$ Now , we will subtract equation (i) by equation (ii). We get... $\\ \to \sf \: - 8 - 4 = a + 8d - (a + 2d) \\$ $\\ \to \sf \: - 12 = a + 8d - a - 2d \\$ $\\ \to\sf \: - 12 = 6d \\$ $\\ \to \boxed{\sf \: d = - 2} \\$ Now , we will substitute the value of "d" in equation (i). $\\ \underline{ \sf \: 4 = a + 2d} \\$ $\\ \to \sf \: 4 = a + 2( - 2) \\$ $\\ \to \sf \: 4 = a - 4 \\$ $\\ \to\boxed{ \sf \:a = 8 } \\ \\$ Let a(n) of the A.P be 0. $\\ \implies\sf \: 0 = a + (n - 1)d \\$ Substituting values of 'a' and 'b' as 8 and -2 RESPECTIVELY , we get... $\\ \implies \sf0 = 8 + (n - 1)( - 2) \\$ $\\ \implies\sf \: - 8 = (n - 1)( - 2) \\$ $\\ \implies \sf - 8 = - 2n + 2 \\$ $\\ \implies\sf2n = 2 + 8 \\$ $\\ \implies\sf 2n = 10 \\$ $\\ \implies \boxed{\boxed{ \bf \:n = 5 }} \\ \\$ HENCE , 5th term of the A.P is 0.