I start with 100 and subtract 3 repeatedly. (100 - 3, 97 - 3, 94 - 3 and so on). How many subtractions would I have to do to get the FIRST single-digit number?
Answers
33.3 i think
but it is easier if you do division
- I would need 32 subtractions in order to get the FIRST single - digit number .
Given :- I start with 100 and subtract 3 repeatedly. (100 - 3, 97 - 3, 94 - 3 and so on) .
To Find :- How many subtractions would I have to do to get the FIRST single - digit number ?
Formula used :- A sequence is said to be in AP (Arithmetic Progression), if the difference between its consecutive terms are equal.
• The nth term of an AP is given as :-
- T(n) = a + (n - 1)•d , where a is the first term and d is the common difference.
• The common difference of an AP is given as :-
- d = T(n) - T(n - 1)
Solution :-
given series is :- 100, 97, 94 _______
As we can see that,
→ 97 - 100 = 94 - 100 = (-3)
since common difference is same, we can conclude that, the given series is in AP with first term as 100 and common difference as (-3) .
So,
→ a = 100
→ d = (-3)
→ T(n) = first single digit number { 9, 8, 7 etc. }
→ T(n) < 10
putting values in above told nth term formula we get,
→ a + (n - 1)d = T(n)
→ 100 + (n - 1)(-3) < 10
→ 100 + (-3n + 3) < 10
→ 100 + 3 - 3n < 10
→ 103 - 3n < 10
→ 103 - 10 < 3n
→ 93 < 3n
dividing both sides by 3,
→ 31 < n .
therefore,
→ n = 32 (Ans.)
Verification :-
putting n = 32 in T(n) we get,
→ T(n) = a + (n - 1)d
→ T(n) = 100 + (32 - 1)(-3)
→ T(n) = 100 + 31 × (-3)
→ T(n) = 100 - 93
→ T(n) = 7 = First single digit number .
Hence, required answer is equal to 32 .
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