Advertisements
Advertisements
Question
Write the first three terms of the APs when a and d are as given below:
a = `sqrt(2)`, d = `1/sqrt(2)`
Solution
Given that,
First term (a) = `sqrt(2)`
and common difference (d) = `1/sqrt(2)`
∵ nth term of an AP,
Tn = a + (n – 1)d
∵ Second term of an AP,
T2 = a + d
= `sqrt(2) + 1/sqrt(2)`
= `(2 + 1)/sqrt(2)`
= `3/sqrt(2)`
and third term of an AP,
T3 = a + 2d
= `sqrt(2) + 2/sqrt(2)`
= `(2 + 2)/sqrt(2)`
= `4/sqrt(2)`
Hence, required three terms are `sqrt(2), 3/sqrt(2), 4/sqrt(2)`.
APPEARS IN
RELATED QUESTIONS
Write the first five terms of the following sequences whose nth terms are:
an = n2 − n + 1
Find the next five terms of the following sequences given by:
`a_1 = -1, a_n = (a_n - 1)/n, n>= 2`
Show that each of the progressions given below is an AP. Find the first term, common difference and next term of each.
(i) 9, 15, 21, 27,…………
Find the 37th term of the AP 6 , 7 `3/4 , 9 1/2 , 11 1/4,.............`
Divide 56 in four parts in AP such that the ratio of the product of their extremes (1st and 4th) to the product of means (2nd and 3rd) is 5 : 6 ?
In an AP, if a = 3.5, d = 0, n = 101, then an will be ______.
If the common difference of an AP is 5, then what is a18 – a13?
The 10th term from the end of the A.P. -5, -10, -15,…, -1000 is ______.
The 10th term from the end of the A.P. 4, 9,14, …, 254 is ______.
Two APs have the same common difference. The first term of one of these is –1 and that of the other is – 8. Then the difference between their 4th terms is ______.
