Advertisements
Advertisements
Question
Find the value of x for which the numbers (5x + 2), (4x - 1) and (x + 2) are in AP.
Advertisements
Solution
It is given that (5x + 2),(4x -1) and (x+2) are in AP.
∴ (4x -1)- (5x +2) = (x+2) - (4x-1)
⇒ 4x -1 - 5x -2 = x+2 -4x +1
⇒ -x -3 = -3x+3
⇒ 3x -x=3+3
⇒2x=6
⇒ x =3
Hence, the value of x is 3.
APPEARS IN
RELATED QUESTIONS
Find the sum of all numbers from 50 to 350 which are divisible by 6. Hence find the 15th term of that A.P.
A ladder has rungs 25 cm apart. (See figure). The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs are 2 `1/2` m apart, what is the length of the wood required for the rungs?
[Hint: number of rungs = `250/25+ 1`]

If the pth term of an A. P. is `1/q` and qth term is `1/p`, prove that the sum of first pq terms of the A. P. is `((pq+1)/2)`.
Determine the A.P. Whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.
Find the sum of first 51 terms of an A.P. whose 2nd and 3rd terms are 14 and 18 respectively.
The first term of an A.P. is 5, the last term is 45 and the sum of its terms is 1000. Find the number of terms and the common difference of the A.P.
The sum of three numbers in AP is 3 and their product is -35. Find the numbers.
The first term of an AP is p and its common difference is q. Find its 10th term.
If the sum of first p terms of an AP is 2 (ap2 + bp), find its common difference.
Find the 25th term of the AP \[- 5, \frac{- 5}{2}, 0, \frac{5}{2}, . . .\]
The next term of the A.P. \[\sqrt{7}, \sqrt{28}, \sqrt{63}\] is ______.
In an A.P. the first term is – 5 and the last term is 45. If the sum of all numbers in the A.P. is 120, then how many terms are there? What is the common difference?
Find the sum of the first 15 terms of each of the following sequences having nth term as xn = 6 − n .
In an A.P. the first term is 8, nth term is 33 and the sum to first n terms is 123. Find n and d, the common differences.
Let Sn denote the sum of n terms of an A.P. whose first term is a. If the common difference d is given by d = Sn − kSn−1 + Sn−2, then k =
The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 is
If \[\frac{5 + 9 + 13 + . . . \text{ to n terms} }{7 + 9 + 11 + . . . \text{ to (n + 1) terms}} = \frac{17}{16},\] then n =
The sum of n terms of an A.P. is 3n2 + 5n, then 164 is its
Obtain the sum of the first 56 terms of an A.P. whose 18th and 39th terms are 52 and 148 respectively.
Find the sum of first 16 terms of the A.P. whose nth term is given by an = 5n – 3.
