Question

how many terms of the AP :9,17,25,.. must be taken to give a sum of 636​

Answer 1

Step-by-step explanation:

AP: 9,17,25......

=> d = 17-9 = 25-17 = 8.

Let the sum of 'n terms' of the AP be equal to 636.

=> Sn = 636

or 636 = n/2{2a + (n-1)d}

or 636 = n/2{18 + (n-1)8}

or 1272 = n(18+8n-8)

or 1272 = n(10+8n)

or 1272 = 8n^2+10n

or 636 = 4n^2+5n

or 4n^2+5n-636 = 0

n = -13.25 (inadmissible) , 12

Therefore, n = 12.

Answer: Hence, 12 terms of the given AP will be required.

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Answer 2

Answer:

12

Step-by-step explanation:

last term of an A.P = a + (n-1)d

where's,

a = first term

n = total term

d = common difference

sum of terms of an A.P = n/2 (a + last term)

= n/2 (a + a + (n-1)d)

= n/2 (2a + dn-d)

= n/2 (2*9 + 8n-8)

= n/2 (18-8+8n)

= n/2 (10+8n)

= (10n+8n^2) / 2

= 636

so,

(10n+8n^2) / 2 = 636

(10n+8n^2) = 636*2=1272

8n^2+10n-1272 = 0

2(4n^2+5n-636) = 0

4n^2+5n-636 = 0/2=0

(4n+53) (n-12) = 0

n = -53/4 (not possible)

n = 12