Sum of the first three terms in an AP is 39 and the product of first and third term is 144. Then, the common difference of the AP is ___ (common difference is greater than 0).

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Solution

Let a-d, a, a+d are the first three terms of an AP.
Sum of the first three terms = 39
a - d + a + a - d = 39
a = 13
Product of first and third term = 144.
(a - d) (a + d) = 144
$a^{2} - d^{2} = 144$
$d^{2} = 25$
d = 5

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Q. The sum of first three terms of an AP is 48. If the product of first and second terms exceeds 4 times the third term by 12, find the AP. [CBSE 2013C]

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Sum of the first three terms in an AP is 39 and the product of first and third term is 144. Then, the common difference of the AP is ___ (common difference is greater than 0).

Let a-d, a, a+d are the first three terms of an AP. Sum of the first three terms = 39 a - d + a + a - d = 39 a = 13 Product of first and third term = 144. (a - d) (a + d) = 144 a2−d2=144 d2=25 d = 5

Let a-d, a, a+d are the first three terms of an AP.
Sum of the first three terms = 39
a - d + a + a - d = 39
a = 13
Product of first and third term = 144.
(a - d) (a + d) = 144
$a^{2} - d^{2} = 144$
$d^{2} = 25$
d = 5