Advertisements
Advertisements
Question
Divide 24 in three parts such that they are in AP and their product is 440.
Solution
Let the required parts of 24 be (a- d) , a and (a +d) such that they are in AP.
Then (a-d) + a+ (a +d) = 24
⇒ 3a = 24
⇒ a=8
Also , (a-d) .a. (a+d) = 440
⇒ `a(a^2 - d^2 )= 440`
⇒` 8(64 - d^2 ) = 440`
⇒`d^2 = 64 - 55 = 9`
⇒ `d= +-3`
Thus , a = 8 and d = `+-3`
Hence, the required parts of 24 are (5,8,11) or (11,8,5).
APPEARS IN
RELATED QUESTIONS
Find the sum of first 15 multiples of 8.
Find the sum 3 + 11 + 19 + ... + 803
Is 184 a term of the AP 3, 7, 11, 15, ….?
The 4th term of an AP is zero. Prove that its 25th term is triple its 11th term.
If 4 times the 4th term of an AP is equal to 18 times its 18th term then find its 22nd term.
If the sum of first m terms of an AP is ( 2m2 + 3m) then what is its second term?
Show that a1, a2, a3, … form an A.P. where an is defined as an = 3 + 4n. Also find the sum of first 15 terms.
The famous mathematician associated with finding the sum of the first 100 natural numbers is ______.
The first term of an AP is –5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference.
Find the sum of first 25 terms of the A.P. whose nth term is given by an = 5 + 6n. Also, find the ratio of 20th term to 45th term.
