In an AP if a = 1, aₙ = 20 and Sₙ = 399, then n is
a. 19
b. 21
c. 38
d. 42
Solution:
An arithmetic progression (AP) is a sequence where the two consecutive terms have the same common difference. It is obtained by adding the same fixed number to its previous term.
From the question,
a = 1
aₙ = 20
Sₙ = 399.
The formula to find the sum is
Sₙ = n/2 [2a + (n - 1)]d.
Substituting the values, we get,
399 = n/2 [2 × 1 + (n - 1)d].
798 = 2n + n(n - 1)d---------------(1)
Since, aₙ = a + (n - 1)d
a + (n-1)d = 20
1 + (n - 1)d = 20
(n - 1)d = 19-------------------------(2)
Substituting (2) in(1), we get,
798 = 2n + 19 n
798 = 21n
n = 798/21.
Therefore, n = 798/21 = 38
✦ Try This: Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the nth week, her weekly savings become Rs 20.75, find n
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 5
NCERT Exemplar Class 10 Maths Exercise 5.1 Problem 17
In an AP if a = 1, aₙ = 20 and Sₙ = 399, then n is, a. 19, b. 21, c. 38, d. 42
Summary:
An arithmetic progression (AP) is a sequence where the two consecutive terms have the same common difference. In an AP if a = 1, aₙ = 20 and Sₙ = 399, then n is 38
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