If the sum of the seriesn 20 + 19(3 / 5) + 19(1 / 5) + 18(4 / 5) + ... up ton n^th term is 488 and then n^th term is negative, then

(b) : Here a = 20,d = (- 2 / 5)n nnIt is given that, S_n = 488 n => (n / 2)lbrack 2 x 20 + (n - 1)( (- 2 / 5) ) rbrack = 488 => (n / 5)lbrack 100 - n + 1rbrack = 488n => n^2 - 101n + 2440 = 0 => n = 61,40n Also, we have,nT_n 102 => n > 51n So,nn = 61Therefore T_n = 20 + 60 x ( (- 2 / 5) ) = 20 - 24 = - 4n

If the sum of the series 20 + 19\frac{3}{5} + 19\fra... | Sequence & Series (Mathematics)

If the sum of the series\n

20 + 19\frac{3}{5} + 19\frac{1}{5} + 18\frac{4}{5} + \ldots

up to\n

n^{\text{th}\text{~}}

term is 488 and the\n

n^{\text{th}\text{~}}

term is negative, then

n = 60

n^{\text{th}\text{~}}

term is -4

n = 41

n^{\text{th}\text{~}}

term is

- 4\frac{2}{5}

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Correct Answer:

n^{\text{th}\text{~}}

term is -4

Explanation

(b) : Here

a = 20,d = \frac{- 2}{5}

\n\end{enumerate}\n\nIt is given that,

S_{n} = 488

\\\n

\Rightarrow \frac{n}{2}\left\lbrack 2 \times 20 + (n - 1)\left( \frac{- 2}{5} \right) \right\rbrack = 488 \Rightarrow \frac{n}{5}\lbrack 100 - n + 1\rbrack = 488\n

\Rightarrow n^{2} - 101n + 2440 = 0 \Rightarrow n = 61,40\n

Also, we have,\n

T_{n} < 0 \Rightarrow 20 + (n - 1)\left( \frac{- 2}{5} \right) < 0

\\\n

\Rightarrow 100 - 2n + 2 < 0 \Rightarrow 2n > 102 \Rightarrow n > 51\n

So,\n

n = 61\therefore T_{n} = 20 + 60 \times \left( \frac{- 2}{5} \right) = 20 - 24 = - 4

Explanation

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