Question

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) 36

Aligarh Muslim University +2 Science/Diploma Entrance Test - 2018-19

Answer 1

Hey friend, Harish here.

Here is your answer:

Given:

1) An series in AP.  : $\frac{5}{6}, \frac{2}{3} , \frac{1}{2}......$

2) Sum of n numbers is $\frac{-121}{2}$

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,

⇒ $S_{n}= \frac{n}{2}(2a + (n-1)d)$

⇒ $\frac{-121}{2} = \frac{n}{2}(2( \frac{5}{6}) + (n-1)( \frac{-1}{6}))$

⇒ $\frac{-121}{2} = (( \frac{5n}{6}) + (n)(n-1)( \frac{-1}{12}))$

⇒ $\frac{-121}{2} =( \frac{(10n + [n(n-1)(-1)])}{12})$

⇒ $-121 \times 6 = 10n + (n^{2}-n)(-1)$

⇒ $-726 = 10n -n^{2} + n$

⇒ $n^{2} -11n - 726 = 0$

⇒ $n^{2} +22n -33n -726 =0$

⇒ $n(n+22)-33(n+22) =0$

⇒ $(n+22) (n-33) =0$

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.

Answer 2

Hi...☺

Here is your answer...✌

Given series :

5/6 , 2/3 , 1/2 ,.....

Here

$a_{2} - a_{1} = \frac{2}{3} - \frac{5}{6} = \frac{4 - 5}{6} = - \frac{1}{6} \\ and \\ a_{3} - a_{2} = \frac{1}{2} - \frac{2}{3} = \frac{3 - 4}{6} = - \frac{1}{6}$

Since, a2-a1 = a3-a2 = -1/6

Therefore given series is an AP

Where first term , a = 5/6
And common diff. , d = -1/6

Now
Given that,

$S _{n} = - \frac{121}{2} \\ \\ \frac{n}{2} [ 2a +( n - 1)d ] = - \frac{121}{2} \\ \\ n [2 \times \frac{5}{6} +( n - 1) \times \frac{ - 1}{6} ] = - 121 \\ \\ n[ \frac{10}{6} - \frac{n}{6} + \frac{1}{6} ] = - 121 \\ \\ n \times \frac{(10 - n + 1)}{6} = - 121 \\ \\ n(11 - n) = - 121 \times 6 \\ \\ 11n - {n}^{2} =- 11 \times 11 \times 3 \times 2 \\ \\ 11n - {n}^{2} + 33 \times 22 = 0 \\ \\ = > {n}^{2} - 11n - 33 \times 22 = 0 \\ \\ n {}^{2} - 33n + 22n - 33 \times 22 = 0 \\ \\ n(n - 33) + 22(n - 33) = 0 \\ \\( n - 33)(n + 22) = 0 \\ \\ = > n - 33 = 0 \: \: or \: \: n + 22 = 0 \\ \\ = > n = 33 \: \: or \: \: n = - 22 \\ \\ = > n = 33 \\ \\ ( by \: rejecting \: n = - 22) \\( \because n \: cannot \: be \: negative)$

HENCE,

OPTION (a) 33 is Correct