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Question
The sum of the first three numbers in an Arithmetic Progression is 18. If the product of the first and the third term is 5 times the common difference, find the three numbers.
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Solution
Let the first three terms be a-d, a, a+d
We have been given that the sum of the first three terms of an A.P is 18
Equation becomes
a - d + a + a + d = 18
3a = 18
⇒ a = 6
Also, we have given the product of first and third term is 5 times the common difference
(a - d) (a + d) = 5d
a2 - d2 = 5d
⇒ a2 = 5d + d2 ................(∵ a = 6)
⇒ d2 + 5d = 36
⇒ d2 + 5d - 36 = 0
d2 + 9d - 4d - 36 = 0
⇒ d (d + 9) - 4 (d + 9) =0
⇒ (d - 4) (d + 9) = 0
⇒ d = 4, -9
When d = 4
First three numbers will be 6 -4, 6, 6+4
⇒ 2, 6, 10
When d= - 9
First three numbers will be 6 - (-9), 6, 6+ (-9)
⇒ 15, 6, -3
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RELATED QUESTIONS
Find the sum of the first 15 terms of each of the following sequences having the nth term as
bn = 5 + 2n
In an A.P. the first term is 25, nth term is –17 and the sum of n terms is 132. Find n and the common difference.
The A.P. in which 4th term is –15 and 9th term is –30. Find the sum of the first 10 numbers.
Find the sum (−5) + (−8)+ (−11) + ... + (−230) .
Write the common difference of an A.P. whose nth term is an = 3n + 7.
In an AP. Sp = q, Sq = p and Sr denotes the sum of first r terms. Then, Sp+q is equal to
Q.15
Q.20
Find the sum of natural numbers between 1 to 140, which are divisible by 4.
Activity: Natural numbers between 1 to 140 divisible by 4 are, 4, 8, 12, 16,......, 136
Here d = 4, therefore this sequence is an A.P.
a = 4, d = 4, tn = 136, Sn = ?
tn = a + (n – 1)d
`square` = 4 + (n – 1) × 4
`square` = (n – 1) × 4
n = `square`
Now,
Sn = `"n"/2["a" + "t"_"n"]`
Sn = 17 × `square`
Sn = `square`
Therefore, the sum of natural numbers between 1 to 140, which are divisible by 4 is `square`.
Which term of the Arithmetic Progression (A.P.) 15, 30, 45, 60...... is 300?
Hence find the sum of all the terms of the Arithmetic Progression (A.P.)
