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8. Sequences and Series
easy
If the sum of three numbers of a arithmetic sequence is $15$ and the sum of their squares is $83$, then the numbers are
A
$4, 5, 6$
B
$3, 5, 7$
C
$1, 5, 9$
D
$2, 5, 8$
Solution
(b) Let three numbers are $a – d,\;a,\;a + d$.
We get $a – d + a + a + d = 15$
$ \Rightarrow $$a = 5$
and ${(a – d)^2} + {a^2} + {(a + d)^2} = 83$
$ \Rightarrow $ ${a^2} + {d^2} – 2ad + {a^2} + {a^2} + {d^2} + 2ad = 83$
$ \Rightarrow $$2({a^2} + {d^2}) + {a^2} = 83$
Putting $a = 5$
$ \Rightarrow $$2(25 + {d^2}) + 25 = 83$
$ \Rightarrow $$2{d^2} = 8$
$ \Rightarrow $$d = 2$
Thus numbers are $3, 5, 7.$
Trick : Since $3 + 5 + 7 = 15$
and ${3^2} + {5^2} + {7^2} = 83$.
Standard 11
Mathematics
