Question

the sum of three consecutive terms of an ap is 21 and the sum of the squares of these terms is 165 find these terms

Answer 1

a-d, a, a+d are in ap
sum=21
a-d +a+a+d=21
3a=21
a=7
(a-d) ^2 + a^2 +(a+d)^2=165
a^2 +d^2-2ad + a^2 + a^2+d^2 + 2ad=165
3a^2+2d^2=165
3(7)^2 + 2d^2=165
3(49) +2d^2=165
2d^2=165-147
2d^2=18
d^2=9
d=3
a-d =7-3=4
a=7
a+d = 7+3=10
numbers are 4,7,10

Answer 2

HEYA!!
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◀ARITHMETIC PROGRESSIONS▶
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➡Let the three Consecutive terms be :

a - d , a and a + d respectively

⚫ACCORDING TO THE GIVEN QUESTION :
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a - d + a + a + d = 21

3a = 21

a = 7
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Thus, the terms become 7-d , 7 and 7+d

Applying second condition ,

(7-d)^2 + (7)^2 + (7+d)^2 = 165

49 + d^2 - 14d + 49 + 49 + d^2 + 14d = 165

147 + 2d^2 = 165

2d^2 = 18

d^2 = 9

d = 3 , -3

➡Taking positive value of d = 3

The terms become (7- 3 ) , 7 and ( 7+3 )

= 4 , 7 and 10


➡Taking negative value of d= -3

The terms become ( 7+3 ) , 7 and (7-3 )

= 10 , 7 and 4

[ Only the order of terms will change ]


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