Advertisements
Advertisements
प्रश्न
Find the 18th term of the AP `sqrt2, 3sqrt2, 5sqrt2.....`
उत्तर
In this problem, we are given different A.P. and we need to find the required term of that A.P.
18th term of the AP `sqrt2, 3sqrt2, 5sqrt2.....`
Here
First term (a) =`sqrt2`
Common difference of the A.P. (d) = `3sqrt2 - sqrt2`
`=2sqrt2`
Now, as we know,
`a_n a + (n -1)d`
So, for 18th term,
`a_18 = a + (18 - 1)d`
`= sqrt2 + (17)2sqrt2`
`= sqrt2 + 34sqrt2`
`= 35sqrt2`
Therefore, the 18th term of the given A.P. is `a_18 = 35sqrt2`
APPEARS IN
संबंधित प्रश्न
The 11th term and the 21st term of an A.P. are 16 and 29 respectively, then find:
(a) The first term and common difference
(b) The 34th term
(c) ‘n’ such that tn = 55
The 16th term of an A.P. is five times its third term. If its 10th term is 41, then find the sum of its first fifteen terms.
Following are APs or not? If they form an A.P. find the common difference d and write three more terms:
0.2, 0.22, 0.222, 0.2222 ….
Find 11th term of the A.P. 10.0, 10.5, 11.0, 11.5, ...
If an A.P. consists of n terms with first term a and nth term l show that the sum of the mth term from the beginning and the mth term from the end is (a + l).
There is an auditorium with 27 rows of seats. There are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row and so on. Find the number of seats in the 15th row and also find how many total seats are there in the auditorium?
Find d if t9 = 23 व a = 7
Common difference, d = ? for the given A.P., 7, 14, 21, 28 ........
Activity :- Here t1 = 7, t2 = 14, t3 = 21, t4 = `square`
t2 − t1 = `square`
t3 – t2 = 7
t4 – t3 = `square`
Therefore, common difference d = `square`
In which of the following situations, do the lists of numbers involved form an AP? Give reasons for your answers
The amount of money in the account of Varun at the end of every year when Rs 1000 is deposited at simple interest of 10% per annum.
First term and common difference of an A.P. are 1 and 2 respectively; find S10.
Given: a = 1, d = 2
find S10 = ?
Sn = `n/2 [2a + (n - 1) square]`
S10 = `10/2 [2 xx 1 + (10 - 1)square]`
= `10/2[2 + 9 xx square]`
= `10/2 [2 + square]`
= `10/2 xx square = square`
