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