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how many terms of an AP 1,4,7...must be taken to give a sum of 51? Expert Answer
Hint
To find 'n' = number of terms.

Given:
AP as 1, 4, 7, ...,
Sn = 51.

a = 1
d = 3 (4 - 1 = 3, 7 - 4 = 3).

Sn = (n/2)[2a + (n - 1)d]
51 = (n/2)[2 + 3n - 3]
102 = n(3n - 1)
so,
3n2 - n - 102 = 0
3n2 - 18n + 17n - 102 = 0
3n(n - 6) + 17(n - 6) = 0
(n - 6)(3n + 17) = 0

Solving this gives us n = 6 and a negative number.
A total of six terms must be taken from the AP to obtain 51.

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