Question

Sharmila needs Rs 1000 for her school trip. She decides to save money for that. On the first day she saves Rs 10. On each succeeding day, she saves Rs 10 more than the money she saved on the previous day. On which day would her total savings just cross Rs 1000?​

Answer 1

Answer:

After 14 days she will have enough money for her field trip

Step-by-step explanation:

10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 + 100 + 110 + 120 + 130 + 140

= 1050Rs

Answer 2

Answer:

On 14th day Sharmila's savings will cross 1000    

Step-by-step explanation:

Given data

Sharmila needs Rs 1000, so that she is saving money as said below

on first day she saves Rs.10. On each succeeding day she saves Rs.10 more than the previous day

here we need to find on which day her savings will cross Rs.1000

⇒ from given data her savings = 10, 20, 30, 40....  which are in A.P

⇒ here first term a = 10 and common difference d = 10

⇒ let's consider that on  n th the day Sharmila's savings will cross 1000

⇒ therefore sum of the n terms

               S = n/2 [2a + (n − 1) × d] $\geq$  1000

                 ⇒  n/2[ 2(10) + (n-1) 10] $\geq$ 1000

                 ⇒ n/2 [ 20 +10n -10 ] $\geq$ 1000

                 ⇒ [n/2 (10 +10n )] 2 $\geq$ 1000 × 2   [multiplied by 2 on both sides]

                 ⇒ n ( 10 +10n ) $\geq$ 2000

                 ⇒ 10n + 10n² $\geq$ 2000

                 ⇒ n + n² $\geq$  200

                 ⇒ n( 1 + n) $\geq$  200 _ (1)   [find for which value (1) will satisfy]

here for n = 14 ⇒ (1) =  14 (15) $\geq$ 200

                                 =  210 $\geq$ 200  

⇒ on 14th day Sharmila's savings will cross 1000