In an A-P- a1- a2- a3- - an- - - if a1-1- an-20 and S n-399- then the value of n is

Given: First term (a) = a1 nbsp;= 1 nth term nbsp;= an nbsp;= 20 Sum of n terms = nbsp;Sn nbsp;= 399 Now, an=a+n 1d #8658;20=1+n 1d #8658;n 1d=20 ...

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Standard X

Mathematics

nth Term of an AP

In an A.P. a1...

Question

In an A.P. a₁, a₂, a₃, ..., a_n, ...., if a₁ = 1, a_n= 20 and S_n= 399, then the value of n is __________.

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Solution

Given:
First term (a) = a₁ = 1
n^th term = a_n= 20
Sum of n terms = S_n= 399

Now,
$a_{n} = a + (n - 1) d \Rightarrow 20 = 1 + (n - 1) d \Rightarrow (n - 1) d = 20 - 1 \Rightarrow (n - 1) d = 19 . . . (1) S_{n} = \frac{n}{2} [2 a + (n - 1) d] \Rightarrow 399 = \frac{n}{2} [2 (1) + 19] (From (1)) \Rightarrow 399 = \frac{n}{2} [21] \Rightarrow 798 = 21 n \Rightarrow n = \frac{798}{21} \Rightarrow n = 38$

Hence, the value of n is 38.

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In an A.P. a1, a2, a3, ..., an, ...., if a1 = 1, an = 20 and Sn = 399, then the value of n is __________.

Given: First term (a) = a1 = 1 nth term = an = 20 Sum of n terms = Sn = 399 Now, an=a+n-1d⇒20=1+n-1d⇒n-1d=20-1⇒n-1d=19 ...1 Sn=n22a+n-1d⇒399=n221+19 From 1⇒399=n221⇒798=21n⇒n=79821⇒n=38 Hence, the value of n is 38.

In an A.P. a₁, a₂, a₃, ..., a_n, ...., if a₁ = 1, a_n= 20 and S_n= 399, then the value of n is __________.