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Which term of the AP : 121, 117, 113, . . ., is its first negative term? [Hint : Find n for < 0]
Given A.P. is 121, 117, 113 …
a = 121
d = 117 − 121 = −4
= 121 + (n - 1) (-4)
= 121 - 4n + 4
= 125 - 4n
To find the first negative term of this A.P.
Therefore, 32nd term will be the first negative term of this A.P.
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of the first sixteen terms of the AP.
Similarly,
Given that,
(a + 2d) + (a + 6d) = 6
2a + 8d = 6
a + 4d = 3
Also, it is given that
(a + 2d) × (a + 6d) = 8
From equation (i),
Solving for d
,
When
When
When
When
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x. [Hint : ]
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