Question

Anvar saves some amount every month . in first three months he saves ₹ 200, ₹250 and ₹ 300 respectively. in which month will he save ₹ 1000 ?​

Answer 1

Answer:

a+(n-1)×d

1000=200+(n-1)×50

1000=200+50n-50

1000=150+50n

50n=1000-150

n=850/50

n=17

Answer 2

Answer:

Saving in first month ₹200; Saving in second month ₹250;...200,250,300. (this is an A.P)

$\sf \: Here \: A=200, \: d=50, \: Let's \: find \: using \: t_n$

$\sf \: Formula \: and \: then \: find \: s_n$

$\sf \: t_n=a+(n-1)d$

$\sf \implies \: \sf200 + (n - 1) - 50 \\ \\ \sf = 200 + 50n - 50 \\ \\ \sf \: 1000 = 150 + 20n$

$\sf \implies \: 150n + 50n = 1000 \\ \\ \sf \: = 50n = 1000 - 150 \\ \\ \sf \: = 50n = 850 \\ \\ \sf \therefore \: n = 17$

$\red{ \sf \: In \: the \: 17 {}^{th} \: month \: he \: will \: save \: ₹1000}$

Let's find that in 17 month how much total amount is saves.

$\sf \implies \sf \: s_n = \frac{n}{2} \: \: \: \: \: \: \: \: \: \: \bigg[2a + (n - 1)d \bigg] \\ \\ = \sf \: \frac{17}{2} \: \: \bigg[2 \times 200 + (17 - 1) \times 50\bigg] \\ \\ \sf = \frac{17}{2} \: \: \: \bigg[400 + 800\bigg] \\ \\ \sf \: \sf \implies17 \times 600 \\ \\ \sf \pink{= 10200}$

Hence, In 17 months total saving is ₹10200.