Question

Sum of three consecutive terms of an AP is
12 and sum of their sequence is 56 Find the
numbers.​

Answer 1

Answer

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Let a−d,a,a+d be the three consecutive terms of an A.P.

Given: Sum=a−d+a+a+d=18

⇒3a=18 or a=6

Sum of their squares=(6−d)2+62+(6+d)2=140

⇒36+d2−12d+36+36+d2+12d=140

⇒2d2=140−108=32

⇒d2=16

⇒d=±4

The three consecutive terms are 6+4,4,6−4 or 6−4,−4,6+4

10.4.2 or 2,−4,10