A car travels $260$ km distance from a place $A$ to $B$ , at a uniform speed $65$ km/hr passes through all thirteen green traffic signals, $4$ minutes at first signal, $7$ minutes at second signal, $10$ minutes at third signal and so on stops for $40$ minutes at thirteenth signal. How much total time it takes to reach at the place $B$ ? Solve by suitable Mathematical Method.

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Solution

Time sequence forms an A.P
$4, 7, 10......$
$a = 4$
$d = 7 - 4 = 3$
For thirteenth station, time $= 40$
total time is the sum of first thirteen terms of the sequence
$S = \frac{n}{2} {2 a + (n - 1) d} S = \frac{13}{2} {2 (4) + (13 - 1) 3} S = \frac{13}{2} \times (8 + 36) S = 13 \times 22 S = 286$
So, total time is $286$ minutes.

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Time sequence forms an A.P4,7,10......a=4d=7−4=3For thirteenth station, time =40total time is the sum of first thirteen terms of the sequence S=n2{2a+(n−1)d}S=132{2(4)+(13−1)3}S=132×(8+36)S=13×22S=286So, total time is 286 minutes.