Advertisements
Advertisements
Question
The product of four consecutive positive integers is 840. Find the numbers.
Solution
Let the four consecutive positive integers be x ,(x + 1), (x + 2) , (x + 3)
∴ x (x+1) (x+2) (x +3) = 840 ...(given)
∴ x (x + 3) (x + 1) (x + 2) = 840
∴ (x2 + 3x) [x (x+2)+1 (x+2)] = 840
∴ (x2 + 3x) [x2 + 2x + x +2] = 840
∴ (x2 + 3x) (x2 + 3x +2) = 840
Let x2 + 3x = m
∴ m (m + 2) = 840
∴ m2 + 2m = 840
∴ m2 + 2m - 840 = 0 ....(+30 ,-28= -840)
∴ m2 + 30m - 28m - 840 = 0
∴ m (m + 30) - 28(m + 30) = 0
∴ (m + 30) (m - 28) = 0
∴ m + 30 = 0 OR m - 28 = 0
∴ m = -30 m = 28
∵ We need positive integers.
∴ m ≠ -30 ∴ m = 28
Resubstituting m = x2 + 3x
∴ x2 + 3x = 28
∴ x2 + 3x -28 = 0
∴ x2 + 7x - 4x - 28 = 0
∴ x (x + 7) - 4 (x +7) = 0
∴ (x + 7) (x - 4) = 0
∴ x + 7 = 0 OR x - 4 = 0
∴ x = -7 OR x = 4
∵ we need positive integers
∴ x ≠ -7 but x = 4
∴ The four consecutive positive integers are 4, 5, 6 and 7 respectively.
APPEARS IN
RELATED QUESTIONS
The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find:
(a) The first term and common difference
(b) The 34th term
(c) ‘n’ such that tn = 55
Babubhai borrows Rs. 4,000 and agrees to repay with a total interest of Rs. 500 in 10 installments, each installment being less than the preceding installment by Rs. 10. What should be the first and the last installments?
If the pth term of an A.P. is q and the qth term is p, prove that its nth term is (p + q – n)
In the following situation, involved make an arithmetic progression? and why?
The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.
Write first four terms of the A.P. when the first term a and the common differenced are given as follows:
a = -2, d = 0
Write first four terms of the A.P. when the first term a and the common difference d are given as follows:
a = -1.25, d = -0.25
Write the next two terms of A.P. whose first term is 3 and the common difference is 4.
If the 10th term of an A.P. is 52 and the 17th term is 20 more than the 13th term, find the A.P.
The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7: 15. Find the numbers.
For the following arithmetic progressions write the first term a and the common difference d:
−5, −1, 3, 7, ...
Find the common difference and write the next four terms of each of the following arithmetic progressions:
0, −3, −6, −9, ...
Find 8th term of the A.P. 117, 104, 91, 78, ...
The first term and the common difference of an A. P. is 10,000 and
2000 resectively. Find the sum of first 12 terms of the A.P.
How many three digit numbers are divisible by 7?
Check whether the following sequence is in A.P.
a – 3, a – 5, a – 7, …
Check whether the following sequence is in A.P.
`1/2, 1/3, 1/4, 1/5, ...`
Which term of an A.P. 16, 11, 6, 1, ... is – 54?
Find d if t9 = 23 व a = 7
Common difference, d = ? for the given A.P., 7, 14, 21, 28 ........
Activity :- Here t1 = 7, t2 = 14, t3 = 21, t4 = `square`
t2 − t1 = `square`
t3 – t2 = 7
t4 – t3 = `square`
Therefore, common difference d = `square`
Find the first terms and common difference of an A.P. whose t8 = 3 and t12 = 52.
