Advertisements
Advertisements
प्रश्न
Determine the A.P. Whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.
Advertisements
उत्तर
For given A.P
`t_3 = a + 2d = 16` ....(i)
Now
`t_7 - t_5 = 12`
`=> (a + 6d) - (a + 4d) = 12`
⇒ 2d = 12
⇒ d = 6
Substituting the value of din (i) we get
`a + 2 xx 6 = 16`
⇒ a + 12 = 16
`=> a = 4`
Thus the required A.P = a, a + d, a + 2d, a + 3d,..
= 4, 10, 16, 22,...
संबंधित प्रश्न
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, .... as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take `pi = 22/7`)
[Hint: Length of successive semicircles is l1, l2, l3, l4, ... with centres at A, B, A, B, ... respectively.]
If the ratio of the sum of the first n terms of two A.Ps is (7n + 1) : (4n + 27), then find the ratio of their 9th terms.
If (3y – 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value of y.
The sum of three consecutive terms of an AP is 21 and the sum of the squares of these terms is 165. Find these terms.
The sum of first three terms of an AP is 48. If the product of first and second terms exceeds 4 times the third term by 12. Find the AP.
HINT: Let these terms be (a – d), a, (a + d).
How many three-digit natural numbers are divisible by 9?
In an A.P. sum of three consecutive terms is 27 and their product is 504, find the terms.
(Assume that three consecutive terms in A.P. are a – d, a, a + d).
Find four consecutive terms in an A.P. whose sum is 12 and sum of 3rd and 4th term is 14.
(Assume the four consecutive terms in A.P. are a – d, a, a + d, a +2d)
If Sn denote the sum of the first n terms of an A.P. If S2n = 3Sn, then S3n : Sn is equal to
The sum of the first 2n terms of the AP: 2, 5, 8, …. is equal to sum of the first n terms of the AP: 57, 59, 61, … then n is equal to ______.
