Answer :

From the given data we calculate the distance covered for each potato.

So, for first potato, distance = 2 × 5 = 10m

For second, distance = 10 + 2 × 3= 16m

For third, distance = 16 + 2 × 3= 22m

Thus the distance to be covered form an AP. :

10, 16, 22, 28, …………

The total distance to be covered for 15 potatoes is given by S_{15} .

Hence, for the above AP, we have a = 10 and d = 6.

So, we know that, S_{n} = × (2a + (n – 1)d)

⇒ S_{15}= × [2(10) + (15 – 1) × 6]

= × [20 + 14 × 6]

= × [20 + 84]

= × 104

= 15 × 52

= 780m

∴ if 15 potatoes are placed in the race, the total distance covered is 780m.

Now, it is given that, the total distance covered is 1340m and we need to find the no. Of potatoes.

So, let the no. Of potatoes be n, then we take,

⇒ S_{n} = 1340

⇒ 1340 = × (2a + (n – 1)d)

⇒ 1340 = × (2(10) + (n – 1)(6))

⇒ 1340 = × (20 + 6n – 6)

⇒ 1340 = × (14 + 6n)

⇒ 1340 = n × (7 + 3n)

⇒ 3n^{2} + 7n – 1340 = 0

On solving we get that,

⇒ n = or n = 20

as we n cannot be negative, so we have n = 20.

∴ if total distance to be covered is 1340m, then no. Of potatoes placed in the race are 20.

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