find 4 numbers in ap whose sum is 20 and the sum of whose squares is 120
Answers
Answer :
2 , 4 , 6 , 8
Solution :
Let the 4 numbers in AP be ;
(a - 3d) , (a - d) , (a + d) , (a + 3d)
Here ,
It is given that , the sum of the 4 numbers in AP is 20 .
Thus ,
=> (a - 3d) + (a - d) + (a + d) + (a + 3d) = 20
=> 4a = 20
=> a = 20/4
=> a = 5
Also ,
It is given that , the sum of squares of the 4 numbers is 120 .
Thus ,
=> (a - 3d)² + (a - d)² + (a + d)² + (a + 3d)² = 120
=> (a - 3d)² + (a + 3d)² + (a - d)² + (a + d)² = 120
=> 2[a² + (3d)²] + 2[a² + d²] = 120
{ °•° (A+B)² + (A-B)² = 2(A² + B²) }
=> 2(a² + 9d²) + 2(a² + d²) = 120
=> 2(a² + 9d² + a² + d²) = 120
=> 2(2a² + 10d²) = 120
=> 2×2(a² + 5d²) = 120
=> 4(a² + 5d²) = 120
=> a² + 5d² = 120/4
=> a² + 5d² = 30
=> 5² + 5d² = 30
=> 25 + 5d² = 30
=> 5d² = 30 - 25
=> 5d² = 5
=> d² = 5/5
=> d² = 1
=> d = √1
=> d = ± 1
• If d = 1 , then ;
1st no. = a - 3d = 5 - 3•1 = 5 - 3 = 2
2nd no. = a - d = 5 - 1 = 4
3rd no. = a + d = 5 + 1 = 6
4th no. = a + 3d = 5 + 3•1 = 5 + 3 = 8
• If d = 1 , then ;
1st no. = a - 3d = 5 - 3•(-1) = 5 + 3 = 8
2nd no. = a - d = 5 - (-1) = 5 + 1 = 6
3rd no. = a + d = 5 + (-1) = 5 - 1 = 4
4th no. = a + 3d = 5 + 3•(-1) = 5 - 3 = 2
Hence ,
Required numbers in AP are ;
2 , 4 , 6 , 8 .
Answer:
Step-by-step explanation:
The four terms of the AP are
a - 3d , a - d , a +d , a + 3d
According to our problem ,
case1; a - 3d + a - d + a +d + a + 3d = 20
4a = 20
a = 20/4
a = 5 ----(1)
case 2;
(a - 3d)² + (a - d )² + (a +d )²+ (a + 3d )²= 120
a² - 6ad + 9 d² + a² - 2d + d² + a² + 2d + d² + a² + 6ad + 9d² = 120
a² + 9d² + a² + d² + a² + d² + a² + 9d² =120
4a² + 20d² = 120
4(a² + 5d²) = 120
a² + 5 d² =30
substituting eqn (1)
25 + 5d² = 30
5d² = 5
d² = 1
d = 1
now substituting a value and d value in 4 terms
we get
a-3d = 5 - 3 =2
a - d = 5 - 1 =4
a + d = 5 + 1 = 6
a + 3d = 5 + 3 = 8
so the four terms of the AP is 2, 4 , 6 , 8
NOTE :
a - d , a , a+ d ; easiest way to find 3 terms
a - 3d , a - d , a + d , a + 3d ; easiest way to find 4 terms
a - 2d , a - d , a ,a + d , a + 2d ; easiest way to find 5 terms
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