Advertisements
Advertisements
प्रश्न
Find the common difference of an A.P. whose first term is 5 and the sum of first four terms is half the sum of next four terms.
Advertisements
उत्तर
Let a and d be the first term and common difference of A.P. respectively.
Given, a = 5
a1 + a2 + a3 + a4 = `(1)/(2)` (a5 + a6 + a7 + a8)
∴ a + (a + d) + (a + 2d) + (a + 3d)
= `(1)/(2)[(a + 4d) + (a + 5d) + (a + 6d) + (a + 7d)]`
⇒ 2(4a + 6d) = (4a + 22d)
⇒ 2(20 + 6d) = (20 + 22d) ...(∵ a = 5)
⇒ 40 + 12d = 20 + 22d
⇒ 10d = 20
⇒ d = 2
Thus, the common difference of A.P. is 2.
APPEARS IN
संबंधित प्रश्न
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
[Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 + 2 × (5 + 3)]
Find the sum of the first 22 terms of the A.P. : 8, 3, –2, ………
Find the sum of first 12 natural numbers each of which is a multiple of 7.
How many numbers are there between 101 and 999, which are divisible by both 2 and 5?
Determine k so that (3k – 2), (4k – 6) and (k + 2) are three consecutive terms of an AP.
If Sn denotes the sum of the first n terms of an A.P., prove that S30 = 3(S20 − S10)
The given terms are 2k + 1, 3k + 3 and 5k − 1. find AP.
Obtain the sum of the first 56 terms of an A.P. whose 18th and 39th terms are 52 and 148 respectively.
How many terms of the A.P. 24, 21, 18, … must be taken so that the sum is 78? Explain the double answer.
Find the sum of all 11 terms of an A.P. whose 6th term is 30.
