Question

Ramkali saved rupees 5 in the first week of a year and then increased her weekly savings by rupees 1.75. If in the nth week, her weekly savings become rupees 20.75, find n.

Answer 1

$\huge \bf{Solution.}$

Here,

a = ₹ 5

d = ₹ 1.75

an = ₹ 20.75

We know that

an = a+(n–1)d

20.75 = 5+(n–1)(1.75)

(n–1)(1.75) = 20.75–5

(n–1)(1.75) = 15.75

n–1 = 15.75/1.75(15.75 upon 1.75)

=1575/175(1575 upon 175) = 315/35(315 upon 35)

n–1=63/7(63 upon 7)

n–1=9

n = 10

Hence, the required value of n is 10.

Answer 2

αղsաҽɾ 

On the 10th week her savings become 20.75 .( n = 10 )

sօlմԵíօղ

Savings made in 1st week = rs 5

Savings made in 2nd week = 5 +1.75 = 6.75

Savings made in 3rd week = 6.75 + 1.75 = 8.50

So the series we got will :-

5, 6.75 , 8.50 ....

As the difference between each number is same so this is A.P series.

Here, Common difference ( d) = 1.75

first term ( a) = 5

And an = 20.75

As we know that :-

an = a +( n-1 ) d .

Using above mentioned formula we got:-

20.75 = 5 + ( n-1) 1.75

15.75 =( n-1)1.75

$\frac{15.75}{1.75} = n - 1$

9 = n-1

n = 10

So the savings will become rs 20.75 on 10th week .