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.

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Solution

$G i v e n t h a t, a = 5 d = 1.75 a_{n} = 20.75 n = ? a_{n} = a + (n - 1) d 20.75 = 5 + (n - 1) \times 1.75 15.75 = (n - 1) \times 1.75 (n - 1) = \frac{15.75}{1.75} = \frac{1575}{175} = \frac{63}{7} = 9 n - 1 = 9 n = 10 H e n c e, n i s 10.$

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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

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 .

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

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. Poonam 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 which week will her weekly saving 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 saving by Rs. 1.75. If in the nth week, her weekly savings become Rs 20.75, find n.

Given that,a=5d=1.75an=20.75n=?an=a+(n−1)d20.75=5+(n−1)×1.7515.75=(n−1)×1.75(n−1)=15.751.75=1575175=637=9n−1=9n=10Hence, n is 10.

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.

$G i v e n t h a t, a = 5 d = 1.75 a_{n} = 20.75 n = ? a_{n} = a + (n - 1) d 20.75 = 5 + (n - 1) \times 1.75 15.75 = (n - 1) \times 1.75 (n - 1) = \frac{15.75}{1.75} = \frac{1575}{175} = \frac{63}{7} = 9 n - 1 = 9 n = 10 H e n c e, n i s 10.$