# 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.

Sn = 300

To find: n

Now, Sn = (n/2) × [2a + (n - 1)d]

Since, Sn = 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 26th term to 36th 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.

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