Sum of three terms in an A.P.is 15 and sum of the squares of the extreme terms is 58. Find the numbers
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Answer:
The answer is 3,5 and 7.
Step-by-step explanation:
Let ( a - d ) , a , ( a + d ) are three numbers in A.P
For first condition:
1) The sum of three number is = 15
a - d + a + a + d = 15 (-d and +d cancel out)
3a = 15
a = 15/3
a = 5
Now for 2nd condition:
sum of squares of the extreme terms = 58
Apply the whole square formula.
( a - d )² + ( a + d )² = 58
a² + d² - 2ad + a² + d² +2ad (after cancellation taking 2 common)
2( a² + d² ) = 58
( a² + d² ) = 58/2
a² + d² = 29
5² + d² = 29 ( a = 5 )
d² = 29 - 25
d² = 4
d = ± 2
Therefore, Required numbers are
If a = 5 , d = 2
1) a - d = 5 - 2 = 3
a = 5
2) a + d = 5 + 2 = 7
Thus the numbers are 3, 5 and 7.
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