Find the value of the middle term of the A.P -6,-2,2,-.....,58. *

1.	Find the value of the middle term of the A.P -6,-2,2,-.....,58. *
Answer» ong>Answer: $\bigstar{\bold{Middle\:<klux>TERM</klux>=26}}$ Step-by-step explanation: $\Large{\underline{\underline{\bf{Given:}}}}$ A.P is -6, -2 , 2 , .....58 $\Large{\underline{\underline{\bf{To\:<klux>FIND</klux>:}}}}$ The middle term of the A.P $\Large{\underline{\underline{\bf{Solution:}}}}$ → First we have to find the number of terms(N) of the A.P → We know that last term of an A.P is given by, $a_n=a_1+(n-1)\times d$ where a₁ = -6, $a_n$ = 58, d = -2 + 6 = 4 → Substituting the datas we get the value for 58 = -6 + (n - 1) × 4 58 = -6 + 4n - 4 58 = -10 + 4n 4n = 58 + 10 4n = 68 n = 68/4 n = 17 → Hence the number of terms in the given A.P is 17 → Since the number of terms is odd, the middle term would be (n+1)/2 th term → Hence the middle term here would be (17 + 1)/2 th term = 18/2 th term = 9th term. → Now we have to fid the ninth term of the A.P → Ninth term of the A.P is given by a₉ = a₁ + 8d → SUBSTITUTE the datas, a₉ = -6 + 8 × 4 a₉ = -6 + 32 a₉ = 26 → Hence the middle term of the A.P is 26 $\boxed{\bold{Middle\:term=26}}$ $\Large{\underline{\underline{\bf{Notes:}}}}$ → The nth term of an A.P is given by $a_n=a_1+(n-1)\times d$ → Sum of n terms of an A.P is given by $S_n=\dfrac{n}{2}(a_1+a_n)$ $S_n=\dfrac{n}{2}(2a_1+(n-1)\times d)$

Answer»

ong>Answer:

$\bigstar{\bold{Middle\:<klux>TERM</klux>=26}}$