Question

please solve it as soon as possible.answer 45th and 61th question

Answer 1

♥️Hope u like my process♥️
============================
45)

Sum of A.P. =

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

_______________________

So..

Sum of First 10 terms

$= \bf \: \frac{10}{2} (2a + (10 - 1)d) \\ \\ or. \: \: - 150 = 5(2a + 9d) \\ \\ or. \: \: 2a + 9d = - 30$

Sum of 20 terms,

$\bf \: = \frac{20}{2} (2a + (20 - 1)d) \\ \\ = 10(2a + 19d) = 10((2a + 9d) + 10d) \\ \\ = 10( - 30 + 10d) \\ \\ = 100(d - 3)$
__________________________

=> So sum of next 10 terms =

Sum of 20 terms - sum of 10 terms

$= > - 550 = 100(d - 3) - ( - 150) \\ \\ or. \: \: 100(d - 3) = - 550 - 150 = - 700 \\ \\ or. \: \: d - 3 = \frac{ - 700}{100} = - 7 \\ \\ or. \: \: \bf \: d = - 7 + 3 = - 4$
______________________________
We know,

$= > \bf \: 2a + 9d = - 30 \\ \\ or. \: \: 2a + 9( - 4) = - 30 \\ \\ or. \: \: 2a = 36 - 30 = 6 \\ \\ or. \: \bf \: \: a \: = \frac{6}{2} = 3$
__________________________
Thus the required A.P is

=> 3, - 1, -5, -9, -13,- 17......
__________________________

61)

The money required by Ramkali =Rs 1800

Initial saving (a) = Rs 50

Monthly saving (d) = Rs 20

Total saving in a year ( 12 months)

$= \bf \: \frac{n}{2} (2a + (n - 1)d )\\ \\ = \frac{12}{2} (2 \times (50 + (12 - 1)20 ) \\ \\ = (100 + (11 \times 20)) \times 6 \\ \\ = (100 + 220) \times 6 = 320 \times 6 \\ \\ \bf \: = Rs \: \: 1920$
_______________________

So she will be able to send her daughter in school.

__________________________
Hope these are the required answers

Proud to help you