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

n = 10

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n = 12

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n = 16

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n = 20

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Solution

The correct option is A $n = 10$
Given that
$a = 5$
$d = 1.75$
$a_{n} = 20.75$
$a_{n} = a + (n - 1) d$
$20.75 = 5 + (n - 1) 1.75$
$15.75 = (n - 1) 1.75$
$(n - 1) = \frac{15.75}{1.75} = \frac{1575}{175} = \frac{63}{7} = 9$
$∴ n - 1 = 9$
$∴ n = 10$

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Similar questions

Ramkali saved $R s . 5$ in the first week of a year and then increased her weekly saving by $R s . 1.75.$ If in the $n^{t h}$ week, her week, her weekly savings become $R s . 20.75,$ find $n .$

Q. Ramkali saved Rs. 5 in the first week of a year and then increased her weekly saving by Rs. 1.75. If in the

n^{t h}

week, her weekly savings become Rs 20.75, find n.

Ramkali saved $R s . 5$ in the first week of a year and then increased her weekly saving by $R s . 1.75$ . If in the $n^{t h}$ week, her weekly savings become $R s . 20.75$ , find $n$ .

Q. Ramkali saved Rs.

5

in the first week of a year and then increased her weekly savings by Rs.

1.75

. If in the nth week, her savings become Rs.

20.75

, find n

Q. Jasleen saved

R s . 5

in the first week of the year and then increased her weekly savings by

R s . 1.75

each week. In what week will her weekly savings be

R s . 20.75

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

The correct option is A n=10Given thata=5d=1.75an=20.75an=a+(n−1)d20.75=5+(n−1)1.7515.75=(n−1)1.75(n−1)=15.751.75=1575175=637=9∴n−1=9∴n=10

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

The correct option is A $n = 10$
Given that
$a = 5$
$d = 1.75$
$a_{n} = 20.75$
$a_{n} = a + (n - 1) d$
$20.75 = 5 + (n - 1) 1.75$
$15.75 = (n - 1) 1.75$
$(n - 1) = \frac{15.75}{1.75} = \frac{1575}{175} = \frac{63}{7} = 9$
$∴ n - 1 = 9$
$∴ n = 10$