The speed of a bus increases by 2 km after every one hour.
The speed of a bus increases by 2 km after every one hour. If the distance travelling in the first one hour was 35 km. what was the total distance travelled in 12 hours?
Answer/Solution
552 km
Steps/Work
Explanation:
Given that distance travelled in 1st hour = 35 km
and speed of the bus increases by 2 km after every one hour
Hence distance travelled in 2nd hour = 37 km
Hence distance travelled in 3rd hour = 39 km
..
Total Distance Travelled = [35 + 37 + 39 + ... (12 terms)]
This is an Arithmetic Progression(AP) with
first term, a=35, number of terms,n = 12 and common difference, d=2.
The sequence a , (a + d), (a + 2d), (a + 3d), (a + 4d), . . .
is called an Arithmetic Progression(AP)
where a is the first term and d is the common difference of the AP
Sum of the first n terms of an Arithmetic Progression(AP),
Sn = n/2[2a+(n−1)d]
where n = number of terms
Hence, [35+37+39+... (12 terms)]
= S12 = 122[2 × 35 +(12 − 1)2]
= 6[70 + 22] = 6 × 92 = 552
Hence the total distance travelled = 552 km
Answer: Option A
Given that distance travelled in 1st hour = 35 km
and speed of the bus increases by 2 km after every one hour
Hence distance travelled in 2nd hour = 37 km
Hence distance travelled in 3rd hour = 39 km
..
Total Distance Travelled = [35 + 37 + 39 + ... (12 terms)]
This is an Arithmetic Progression(AP) with
first term, a=35, number of terms,n = 12 and common difference, d=2.
The sequence a , (a + d), (a + 2d), (a + 3d), (a + 4d), . . .
is called an Arithmetic Progression(AP)
where a is the first term and d is the common difference of the AP
Sum of the first n terms of an Arithmetic Progression(AP),
Sn = n/2[2a+(n−1)d]
where n = number of terms
Hence, [35+37+39+... (12 terms)]
= S12 = 122[2 × 35 +(12 − 1)2]
= 6[70 + 22] = 6 × 92 = 552
Hence the total distance travelled = 552 km
Answer: Option A