How many terms of the AP ,... must be taken so that their sum is 300? Explain the double answer.

Here, first term = a =20

Common difference = d = 58/3 - 20 = - 2/3

Let first n terms of the AP sums to 300.

∴ S_n = 300

To find: n

Now, S_n = (n/2) × [2a + (n - 1)d]

Since, S_n = 300

∴ (n/2) × [2a + (n - 1)d] = 300

⇒ (n/2) × [2(20) + (n - 1)(-2/3)] = 300

⇒ (n/2) × [40 - (2/3)n + (2/3)] = 300

⇒ (n/2) × [(120 - 2n + 2)/3] = 300

⇒ n[122 - 2n] = 1800

⇒ 122n - 2n² = 1800

⇒ 2n² - 122n + 1800 = 0

⇒ n² - 61n + 900 = 0

⇒ (n - 36)( n - 25)= 0

⇒ n = 36 or n = 25

∴ n= 36 or n = 25

Now, S₃₆ = (36/2)[2a + 35d]

= 18(40 + 35(-2/3))

= 18(120 - 70)/3

= 6(50)

= 300

Also, S₂₅ = (25/2)[2a + 24d]

= (25/2)(40 + 24(-2/3))

= (25/2)(40 - 16)

= (24 × 25)/2

= 12 × 25

= 300

Now, sum of 11 terms from 26^th term to 36^th term = S₃₆ - S₂₅ = 0

∴ Both the first 25 terms and 36 terms give the sum 300 because the sum of last 11 terms is 0. So, the sum doesn’t get affected.