Advertisements
Advertisements
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.
Advertisements
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
APPEARS IN
RELATED QUESTIONS
Find the sum given below:
`7 + 10 1/2 + 14 + ... + 84`
Show that a1, a2,..., an... form an AP where an is defined as below:
an = 9 − 5n
Also, find the sum of the first 15 terms.
Find the sum of first 20 terms of the sequence whose nth term is `a_n = An + B`
The sum of three numbers in AP is 3 and their product is -35. Find the numbers.
Write an A.P. whose first term is a and the common difference is d in the following.
a = 10, d = 5
Find the sum: 1 + 3 + 5 + 7 + ... + 199 .
Find the sum \[7 + 10\frac{1}{2} + 14 + . . . + 84\]
Write the nth term of the \[A . P . \frac{1}{m}, \frac{1 + m}{m}, \frac{1 + 2m}{m}, . . . .\]
Find the sum of first 17 terms of an AP whose 4th and 9th terms are –15 and –30 respectively.
If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.
