the sum of five numbers in A.P. is 15 and the sum of their squares is 55. find the numbers.
step by step explanation.
Answers
Sum of first 5 numbers =52(2a+4d)=25
⇒2a+4d=2×255⇒2a+4d=2×5
⇒a+2d=5⇒a=5–2d … (1)
Sum of squares of first n numbers
=({(2n−1)d26+ad}(n−1)+a2)×n
(In case you don’t know this formula, keep a note of it)
Sum of squares of first 5 numbers
=({(2×5−1)d26+ad}(5−1)+a2)×5=165
⇒({96d2+ad}4+a2)×5=165
⇒({1.5d2+ad}4+a2)×5=165
⇒(6d2+4ad+a2)×5⇒30d2+20ad+5a2=165
Putting value of (1),
⇒30d2+20(5–2d)d+5(5–2d)2=165
⇒30d2+100d–40d2+5(25+4d2−20d)=165
⇒30d2+100d–40d2+125+20d2–100d=165
⇒10d2=40
⇒d2=4⇒d=±2
So,
a=5–2d=5±2×2=5±4=9 or 1
The series can be: 1,3,5,7,9 or 9,7,5,3,1
EXPLANATION.
The sum of 5 numbers in an ap is 15.
The sum of their squares is 55.
As we know that,
Five numbers in an ap.
⇒ (a - 2d), (a - d), (a), (a + d), (a + 2d).
The sum of 5 numbers in an ap is 15.
⇒ (a - 2d) + (a - d) + (a) + (a + d) + (a + 2d) = 15.
⇒ a - 2d + a - d + a + a + d + a + 2d = 15.
⇒ 5a = 15.
⇒ a = 3.
The sum of their squares is 55.
⇒ (a - 2d)² + (a - d)² + (a)² + (a + d)² + (a + 2d)² = 55.
⇒ (a² + 4d² - 4ad) + (a² + d² - 2ad) + a² + (a² + d² + 2ad) + (a² + 4d² + 4ad) = 55.
⇒ a² + 4d² - 4ad + a² + d² - 2ad + a² + a² + d² + 2ad + a² + 4d² + 4ad = 55.
⇒ a² + a² + a² + a² + a² + 4d² + d² + d² + 4d² - 4ad - 2ad + 2ad + 4d = 55.
⇒ a² + a² + a² + a² + a² + 4d² + d² + d² + 4d² = 55.
⇒ 5a² + 10d² = 55.
Put the value of a = 3 in the equation, we get.
⇒ 5(3)² + 10d² = 55.
⇒ 5(9) + 10d² = 55.
⇒ 45 + 10d² = 55.
⇒ 10d² = 55 - 45.
⇒ 10d² = 10.
⇒ d² = 1.
⇒ d = ± √1.
⇒ d = ± 1.
Five numbers in an ap.
⇒ (a - 2d), (a - d), (a), (a + d), (a + 2d).
When : a = 3 and d = 1.
⇒ (a - 2d) = [3 - 2(1)] = 1.
⇒ (a - d) = [3 - 1] = 2.
⇒ (a) = 3.
⇒ (a + d) = [3 + 1] = 4.
⇒ (a + 2d) = [3 + 2(1)] = 5.
∴ The numbers are : 1, 2, 3, 4, 5.
When a = 3 and d = - 1.
⇒ (a - 2d) = [3 - 2(-1)] = 5.
⇒ (a - d) = [3 - (-1)] = 4.
⇒ (a) = 3.
⇒ (a + d) = [3 + (-1)] = 2.
⇒ (a + 2d) = [3 + 2(-1)] = 1.
∴ The numbers are : 5, 4, 3, 2, 1.