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.)
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
Find the sum of all natural numbers between 250 and 1000 which are exactly divisible by 3
If the nth term of the A.P. 9, 7, 5, ... is same as the nth term of the A.P. 15, 12, 9, ... find n.
The A.P. in which 4th term is –15 and 9th term is –30. Find the sum of the first 10 numbers.
Divide 207 in three parts, such that all parts are in A.P. and product of two smaller parts will be 4623.
Find the sum of n terms of the series \[\left( 4 - \frac{1}{n} \right) + \left( 4 - \frac{2}{n} \right) + \left( 4 - \frac{3}{n} \right) + . . . . . . . . . .\]
Which term of the sequence 114, 109, 104, ... is the first negative term?
Write the nth term of the \[A . P . \frac{1}{m}, \frac{1 + m}{m}, \frac{1 + 2m}{m}, . . . .\]
Q.12
Q.17
Kanika was given her pocket money on Jan 1st, 2008. She puts Rs 1 on Day 1, Rs 2 on Day 2, Rs 3 on Day 3, and continued doing so till the end of the month, from this money into her piggy bank. She also spent Rs 204 of her pocket money, and found that at the end of the month she still had Rs 100 with her. How much was her pocket money for the month?
