Question

please answer this question ​

Answer 1

Answer:

15

Step-by-step explanation:

a = 21, d = - 3, Sn = 0, n = ?

We know,

$sn = \frac{n}{2} (2a + (n - 1)d)$

$0 = \frac{n}{2} (42 + (n - 1) - 3)$

0 = n(42 - 3n + 3)

0 = n(45 - 3n)

0 = 45 - 3n

3n = 45

n = 15

Answer 2

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$a = 21$

$d = a1- a2$

$d = - 3$

$s{n} = 0$

$sn = \frac{n}{2} (2a + ( - 1)d) \\ \frac{0 \times 2}{n} = 2 + (n -1)d \\ 2 + (n - 1)d = 0 \\ n - 1 = \frac{ - 2a}{d} \\ n - 1 = \frac{ - 2 \times 21}{ - 3} \\ n = \frac{42}{3} + 1 \\ n = 14 + 1 \\ n = 15$

So, 15 terms must be added to get the sum 0.