Q.1. . The sum of three numbers in an Arithmetic Progression is 45 and their product is 3000. What are the three numbers?
Q.2. .The sum of the first six terms of an AP is 48 and the common difference is 2. What is the 4th term?
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Q3. The sum of three numbers in Arithmetic Progression is 72 and their product is 11880. What are the numbers?
Q4. The first term of an AP is 3 and the last term is 17. If the sum of all terms is 150, what is the 5th term?
Answers
[1]
Given :-
◉ Sum of three numbers is 45 which are in AP and thier product is 3000.
Find :-
◉ Numbers
Solution :-
Let the numbers or the three terms of the AP be,
⇒ a - d , a , a + d
Now, It is given that,
⇒ Sum of terms = 45
⇒ a - d + a + a + d = 45
⇒ 3a = 45
⇒ a = 15
Also, It is given that the product of the terms is 3000,
⇒ Product of terms = 3000
⇒ (a + d)(a - d) a = 3000
⇒ (a² - d²)a = 3000
Substituting value of a = 15
⇒ (15² - d²)15 = 3000
⇒ (225 - d²)15 = 3000
⇒ 3375 - 15d² = 3000
⇒ -15d² = -375
⇒ d² = 25
⇒ d = 5
Now, The three numbers we assumed is,
⇒ a - d , a , a + d
⇒ 15 - 5 , 15 , 15 + 5
⇒ 10 , 15 , 20
Hence, The numbers are 10, 15 and 20.
[2]
Given :-
◉ Sum of first six terms is 48 which are in AP. The common Difference of the given AP is 2.
To Find :-
◉ 4th term.
Solution :-
Let the numbers which are in AP be,
⇒ a - 2d , a - d , a , a + d , a + 2d , a + 3d
Substituting d = 2, as given in the question,
⇒ a - 4 , a - 2 , a , a + 2 , a + 4 , a + 6
Now, It is given that the sum of the terms is 48,
⇒ a - 4 + a - 2 + a + a + 2 + a + 4 + a + 6 = 48
⇒ 6a + 6 = 48
⇒ a + 1 = 8
⇒ a = 7
Now, that we have found out the value of a and d, let us find the 4th term,
⇒ a₄ = a + 3d
⇒ a₄ = 7 + 3×2
⇒ a₄ = 7 + 6
⇒ a₄ = 13
Hence, 4th term of the AP is 13.
[3]
Given :-
◉ Sum of three numbers which are in AP is 72, and the product of these terms is 11880
To Find :-
◉ The numbers
Solution :-
Let the three numbers which are in AP be,
⇒ a - d , a , a + d
Now, It is given that,
⇒ Sum of terms = 72
⇒ a - d + a + a + d = 72
⇒ 3a = 72
⇒ a = 24
Now, The product of these terms is 11880,
⇒ (a - d)(a + d)a = 11880
⇒ (a² - d²)a = 11880
⇒ (24² - d²)24 = 11880
⇒ (576 - d²)24 = 11880
⇒ 13824 - 24d² = 11880
⇒ 24d² = 1944
⇒ d² = 81
⇒ d = ± 9
Now the assumed numbers are,
⇒ 24 ± 9 , 24 , 24 ± 9
⇒ 24 - 9 , 24 , 24 + 9
⇒ 15 , 24 , 33
Because whether you take positive d or negative numbers will be same.
Hence, The numbers are 15, 24 and 33.
[4]
Given :-
◉ An AP, in which the first term and the second term are 3 and 17 respectively.
◉ Sum of all terms of the AP is 150.
To Find :-
◉ 5th term of the AP
Solution :-
It is given that the sum of all terms of the AP is 150, Also
- First term, a = 3
- Last term, l = 17
We know,
⇒ Sum of n terms = n/2 [ a + l ]
⇒ 150 = n/2 [ 17 + 3 ]
⇒ 300 = 20n
⇒ n = 30 / 2
⇒ n = 15
Now, Let us find the common difference,
⇒ a₁₅ = a + (15 - 1)d
⇒ a₁₅ = 3 + 16d
⇒ 17 - 3 = 14d
⇒ 14 = 14d
⇒ d = 1
Now that we have found out the value of d,
⇒ a₅ = a + 4d
⇒ a₅ = 3 + 4×1
⇒ a₅ = 7
Hence, The fifth term of the AP is 7.