A particular A.P. has a positive common difference and is such that for any three adjacent terms, 3
times the sum of their squares exceeds the square of their sum by 37.5. Find the common difference.
(CGCEB).
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Answer:
The common difference is 2.5
Step-by-step explanation:
Let the terms of the AP be a-d, a and a+d
where a is the first term of the AP
and d is the common difference
According to the condition in the question,
3[ (a-d)² + a² + (a+d)² ] = (a-d+a+a+d)² + 37.5
=> 3[a² +d² - 2ad + a² + a² + d² + 2ad] = (3a)² + 37.5
=> 3[3a² + 2d²] = 9a² + 37.5
=> 9a² + 6d² = 9a² + 37.5
=> 6d² = 37.5
=> d² = 6.25
Taking square root on both side, we get,
d = 2.5
Therefore, the common difference is 2.5
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