Question

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Answer 1

Answer:

AP : 55, 51, 47, ....

a = 55

d = 51 - 55 = - 4

Sn = 405

$\frac{n}{2} (2a + (n - 1)d) = 405 \\ \\ \frac{n}{2} (2(55) + (n - 1)( - 4)) = 405 \\ \\ \frac{n}{2} (110 - 4n + 4) = 405 \\ \\ n (114 - 4n) = 810 \\ 2n(57 - 2n) = 810 \\ n(57 - 2n) = 405 \\ 57n - 2 {n}^{2} = 405 \\ 2 {n}^{2} - 57n + 405 = 0 \\ 2 {n}^{2} - 30n - 27n + 405 = 0 \\ 2n(n - 15) - 27(n - 15) = 0 \\ (2n - 27)(n - 15) = 0$

Therefore, values for n are

$n = \frac{27}{2} \: and \: \: n \: = 15$

So there are 15 terms for the given AP whose sum is 405.

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Answer 2

See the attachment !

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