How many terms of the arithmetic series 5/6 + 2/3 + 1/2 ......... must be taken in order to obtain a sum of -121/2?(a) 33 (b) 34 (c) 35 (d) 36Aligarh Muslim University +2 Science/Diploma Entrance Test - 2018-19

How many terms of the arithmetic series 5/6 + 2/3 + 1/2 ......... must

1.	How many terms of the arithmetic series 5/6 + 2/3 + 1/2 ......... must be taken in order to obtain a sum of -121/2?(a) 33 (b) 34 (c) 35 (d) 36Aligarh Muslim University +2 Science/Diploma Entrance Test - 2018-19
Answer» Hey FRIEND, Harish here. Here is your answer: *Given:* $\frac{5}{6}, \frac{2}{3} , \frac{1}{2}......$ 1) An series in AP. : $\frac{-121}{2}$ 2) Sum of n numbers is *To find,* The value of N. *Solution:* Here: $First\ term\ (a) = \frac{5}{6}$ ⇒ $Common\ Difference\ (d)=( \frac{2}{3} - \frac{5}{6} )= (\frac{-4+5}{6}) = \frac{-1}{6}$ ⇒ $S_{n} = \frac{-121}{2}$ ⇒ We know that, $S_{n}= \frac{n}{2}(2a + (n-1)d)$ ⇒ $\frac{-121}{2} = \frac{n}{2}(2( \frac{5}{6}) + (n-1)( \frac{-1}{6}))$ ⇒ $\frac{-121}{2} = (( \frac{5n}{6}) + (n)(n-1)( \frac{-1}{12}))$ ⇒ $\frac{-121}{2} =( \frac{(10n + [n(n-1)(-1)])}{12})$ ⇒ $-121 \times 6 = 10n + (n^{2}-n)(-1)$ ⇒ $-726 = 10n -n^{2} + n$ ⇒ $n^{2} -11n - 726 = <klux>0</klux>$ ⇒ $n^{2} +22n -33n -726 =0$ ⇒ $n(n+<klux>22</klux>)-33(n+22) =0$ ⇒ $(n+22) (n-33) =0$ ⇒ If this equation must be zero Then, i) (n+22) = 0. ⇒ n = -22 (This is not possible, Because the number of terms must always be positive) ii) (n - 33) = 0 ⇒ n = 33. (This is possible). *THEREFORE sum of 33 terms in AP series is -121/2. (OPTION - A).* ________________________________________________________ Hope my answer is helpful to you.