Advertisements
Advertisements
Question
Which term of the Arithmetic Progression (A.P.) 15, 30, 45, 60...... is 300?
Hence find the sum of all the terms of the Arithmetic Progression (A.P.)
Advertisements
Solution
Given A.P. is 15, 30, 45, 60.........
Here a = 15, d = 30 – 15 = 15
Let Tn = 300
Tn = a + (n – 1)d
300 = 15 + (n – 1) × 15
300 = 15 + 15n – 15
15n = 300
∴ n = 20
Hence, 300 is the 20th term
Also by Sn = `n/2 [2a + (n - 1)d]`
S20 = `20/2 [2 xx 15 + (20 - 1) xx 15]`
= 10[30 + 285]
= 10 × 315
∴ S20 = 3150
APPEARS IN
RELATED QUESTIONS
How many multiples of 4 lie between 10 and 250?
Find the sum of first 40 positive integers divisible by 6.
If the pth term of an AP is q and its qth term is p then show that its (p + q)th term is zero.
If the sum of a certain number of terms starting from first term of an A.P. is 25, 22, 19, ..., is 116. Find the last term.
The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is given by \[\frac{l^2 - a^2}{k - (l + a)}\] , then k =
Q.19
The sum of the first three numbers in an Arithmetic Progression is 18. If the product of the first and the third term is 5 times the common difference, find the three numbers.
Write the formula of the sum of first n terms for an A.P.
Find the sum of the integers between 100 and 200 that are
- divisible by 9
- not divisible by 9
[Hint (ii) : These numbers will be : Total numbers – Total numbers divisible by 9]
The sum of n terms of an A.P. is 3n2. The second term of this A.P. is ______.
