Question

390 plants are to be planted in a garden in a number of rows. There are 40 plants in the 1" row, 38 plants in the
second row, 36 plants in 3 row and so on. In how many rows the 390 plants are planted ? Find the no of plants in
the last row also

Answer 1

Answer:

12 plants in the 15th row

Step-by-step explanation:

Total no of plants = 390

No of plants in first row = 40

No of plants in second row= 38

No of plants in third row = 36

So if we will do it consecutively then

40+38+36+34+32+30+28+26+24+22+20+18+16+14+12= 390

So,There are 12 plants planted in the 15th row

Answer 2

Answer:

In 15 rows 390 plants are planted.

12 plants are planted in the last row.

Step-by-step explanation:

A.P = a1, a2, a3, _ _ _ _ an

A.P = 40, 38, 36, _ _ _ _ an

d = 38 - 40 = -2

a = 40

Sum of A.P = 390

Let the no. of rows in which 390 plants are planted be n,

n/2[2a + (n - 1)d] = 390

n/2[2 x 40 + (n - 1)(-2)] = 390

n/2[80 + (n - 1)(-2)] = 390

n[80 -2n + 2] = 390 x 2

n[80 - 2n + 2] = 780

80n - 2n² + 2n = 780

-2n² + 82n - 780 = 0

n² - 41n + 390 = 0

n² - 26n - 15n + 390 = 0

n(n - 26) - 15(n - 26) = 0

(n - 26)(n - 15) = 0

n = 26, 15

If n = 26 then Sn = 728

and if n = 15 then n = 390

So, n = 15

Let the no. of plants in the last row be $a_{15}$,

$a_{15}$ = a + (n - 1)d

$a_{15}$ = 40 + (15 - 1)(-2)

$a_{15}$ = 40 + 14(-2)

$a_{15}$ = 40 - 28

$a_{15}$ = 12

I hope this answer is helpful for you.