Let there be an A.P. with the first term 'a', common difference'. If an a denotes in nth term and Sn the sum of first n terms, find.
n and a, if an = 4, d = 2 and Sn = −14.
Solution
Here, we have an A.P. whose nth term (an), the sum of first n terms (Sn) and common difference (d) are given. We need to find the number of terms (n) and the first term (a).
Here
Last term (`a_n`) = 4
Common difference (d) = 2
Sum of n terms (Sn) = −14
So here we will find the value of n using the formula, `a_n = a + (n - 1)d`
So, substituting the values in the above-mentioned formula
4 = a + (n - 1)2
4 = a + 2n - 2
4 + 2 = a + 2n
`n = (6 - a)/2` ....(1)
Now, here the sum of the n terms is given by the formula,
`S_n = (n/2) (a + l)`
Where, a = the first term
l = the last term
So, for the given A.P, on substituting the values in the formula for the sum of nterms of an A.P., we get,
`-14 = (n/2)[a + 4]`
-14(2) = n(a + 4)
`n = (-28)/(a + 4)` ...(2)
Equating (1) and (2), we get,
`(6 - a)/2 = (-28)/(a + 4)`
(6 - a)(a + 4) = -28(2)
`6a - a^2 + 24 - 4a = -56`
`-a^2 + 2a + 24 + 56 = 0`
So, we get the following quadratic equation,
`-a^2 + 2a + 80 = 0`
`a^2 - 2a - 80= 0`
Further, solving it for a by splitting the middle term,
`a^2 - 2a - 80 = 0`
`a^2 - 10a + 8a - 80 = 0`
a(a - 10) + 8(a - 10) = 0
(a - 10)(a + 8) = 0
So, we get,
a - 10 = 0
a = 10
or
a + 8 = 0
a = -8
Substituting, a = 10 in (1)
`n = (6 - 10)/2`
`n = (-4)/2`
n = -2
Here, we get n as negative, which is not possible. So, we take a = -8
`n = (6 - (-8))/2`
`n = (6 + 8)/2`
`n = 14/2`
n = 7
Therefore, for the given A.P n = 7 and a = -8
