Question

In an AP, given a = 2, d = 8, and S_n = 90, find n and a_n.

Let there be an A.P. with the first term 'a' and the common difference 'd'. If a_n a denotes the n^th term and Sn is the sum of the first n terms, find.

n and a_n, if a = 2, d = 8, and S_n = 90.

Answer 1

Given that a = 2, d = 8, and S_n = 90

`"As"  S_n = n/2 [2a + (n - 1)d]`

`90 = n/2 [2xx2 + (n - 1)8]`

90 × 2 = 4n + n(11 - 1) × 8

180 = 4n + 8n² - 8n

180 = 8n² - 4n

45 = 2n² - n

2n² - n - 45 = 0

2n² - 10n + 9n - 45 = 0

2n (n - 5) + 9(n - 5) = 0

(2n + 9) (n - 5) = 0

∴ Either 2n + 9 = 0

n = `-9/2`

or n - 5 = 0

n = 5

But n = `9/2` is not possible.

∴ n = 5

Now, a_n = a + (n - 1)d

a₅ = 2 + (5 -1) × 8

a₅ = 2 + 32

a₅ = 34

Thus, n = 3 and a₅ = 34

Answer 2

Here, we have an A.P. whose first term (a), the common difference (d) and the sum of the first n terms are given. We need to find the number of terms (n) and the n^th term (a_n).

Here,

First term (a) = 2

The sum of the first n^th terms (`S_n`) = 90

Common difference (d) = 8

So, to find the number of terms (n) of this A.P., we use the following formula for the sum of n terms of an A.P

`S_n = n/2 [2a +  (n - 1)d]`

Where a is the first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

So, using the formula for n = 8, we get,

`S_n = n/2 [2(2) + (n-1)(8)]]`

`90  = n/2 [4 + 8n - 8]`

90(2) = n[8n - 4]

`180 = 8n^2 - 4n`

Further solving the above quadratic equation,

`8n^2 - 4n - 180 = 0`

`2n^2 - n - 45 = 0`

Further solving for n,

`2n^2 - 10n + 9n - 45 = 0`

2n(n - 5) + 9(n - 5) = 0

(2n + 9)(n - 5) = 0

Now

2n + 9 = 0

2n = -9/2

Also

n - 5 = 0

n = 5

Since n cannot be a fraction,

Thus, n = 5

Also, we will find the value of the n^th term (a_n) using the formula `a_n = a + (n - 1)d`

So, substituting the values in the above-mentioned formula

`a_n = 2 + (5 - 1)8`

`a_n = 2 + (4)(8)`

`a_n = 2 + 32`

`a_n = 34`

Therefore, for the given A.P n = 5 and `a_n = 34`