The sum of the first three terms in an A.P is 21 and the sum of the square is 115
Find the numbers
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Answer:
Let the terms are a−d,a,a+d
The sum of three numbers in A.P. is 21
∴sum=(a−d)+a+(a+d)=3a=21
⇒a=7
Second condition:-
The sum of their squares is 197.
∴(a+d)
2
+a
2
+(a−d)
2
=197
a
2
+d
2
−2ad+a
2
+a
2
+d
2
+2ad=197
⇒3a
2
+2d
2
=197
3×7
2
+2d
2
=197
⇒147+2d
2
=197
⇒2d
2
=18
d=3
so the series is-
a−d=7−3=4
a=7
a+d=7+3=10
Hence the series is 4,7,10 and the middle no. is 7.
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