Find the sum of the first $15$ terms of the sequence having $n$ th term as
$y_{n} = 9 - 5 n$

- 390

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468

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- 465

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- 468

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Solution

The correct option is C $- 465$
$y_{n} = 9 - 5 n$

$y_{1} = 9 - 5 = 4$

$y_{2} = 9 - 10 = - 1$

$d = y_{2} - y_{1} = - 1 - 4 = - 5$

Thus, $S_{15} = \frac{15}{2} [2 (4) + 14 (- 5)]$

$= \frac{15}{2} [8 - 70] = - 465$

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Find the sum of the first 15 terms of each of the following sequences having nth term as
(i) $a_{n} = 3 + 4 n$ (ii) $b_{n} = 5 + 2 n$ (iii) $x_{n} = 6 - n$ (iv) $y_{n} = 9 - 5 n$

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Find the sum of the first 15 terms of the sequence having nth term asyn=9−5n

The correct option is C −465yn=9−5ny1=9−5=4y2=9−10=−1d=y2−y1=−1−4=−5Thus, S15=152[2(4)+14(−5)]=152[8−70]=−465

Find the sum of the first $15$ terms of the sequence having $n$ th term as
$y_{n} = 9 - 5 n$

The correct option is C $- 465$
$y_{n} = 9 - 5 n$

$y_{1} = 9 - 5 = 4$

$y_{2} = 9 - 10 = - 1$

$d = y_{2} - y_{1} = - 1 - 4 = - 5$

Thus, $S_{15} = \frac{15}{2} [2 (4) + 14 (- 5)]$

$= \frac{15}{2} [8 - 70] = - 465$