# 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. - Mathematics

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

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#### APPEARS IN

RD Sharma Class 10 Maths
Chapter 5 Arithmetic Progression
Exercise 5.6 | Q 56.2 | Page 53