Question

how many term of the series 1 + 5 + 9 + and so on must be taken so that their sum is 190?​

Answer 1

Answer:n=10 or n=-19/2

Step-by-step explanation:

Sn=190

a=1

d=5-1

=4

Sn=n/2(2a+(n-1)d)

190=n/2(2(1)+(n-1)4)

380=n[2+4n-4]

380=n[-2+2n]

380=-2n+4n^2

n=-19/2 and n=20

Answer 2

Answer:

$\huge \red{ \mathfrak{solution}} \\ \\ \\ we \: have \: \\ \\ 1 + 5 + 9...............n \: \\ given \: sum \: is \: \pink{190} \\ first \: term \: a \: = 1 \\ \\ common \: difference \: is \: 4 \\ \\ we \: know \: that \\ \\ \implies \: sum = \frac{n}{2} (2a \: + nd - d) \\ \\ \implies190 = \red{\frac{n}{2} (2 + 4n - 4)} \\ \\ \implies \: 190 = \green{n(2n - 1) }\\ \\ \implies \: 2 {n}^{2} - n - 190 = 0 \: \\ \\ now \: by \: shree \: dharacharya \: principle \\ \\ n = \frac{1 + \sqrt{1520} }{4} or \: \frac{1 - \sqrt{1520} }{4} \\ \\ n = \frac{1 + 39}{4} or \: \frac{1 - 39}{4} \\ \\ n = 10 \: or \frac{ - 38}{4} \\ here \: \frac{ - 38}{4} does \: not \: exist \\ \\ so \: \\ \: n = 10$