Question

Given: an = 4, d = 2 and Sn = -14. Find n and a.

Answer 1

Hint: The sequence of numbers where the difference of any two consecutive terms is the same is called an Arithmetic Progression.
If a be the first term, d be the common difference and n be the number of terms of an AP, then the sequence can be written as follows:
a, a + d, a + 2d, ..., a + (n - 1)d.
The nth term an = a + (n - 1)d.
The sum of n terms of the above series is given by:
 ${{S}_{n}}=\dfrac{n}{2}[a+\{a+(n-1)d\}]=\left( \dfrac{\text{First Term}+\text{Last Term}}{2} \right)\times n$

Complete step-by-step answer:
Let's say that the first term of the Arithmetic Progression is "a" and the common difference is d = 2.
It is given that the nth term is an = 4.
Using the formula for the nth term, we get:
an = a + (n - 1)d
Substituting the value of an and d, we get:
⇒ 4 = a + (n - 1)2
⇒ a + 2n = 6
⇒ a = 6 - 2n ...(1)
It is also given that the Sn = -14.
Using the formula for the sum of first n terms Sn, we get:
 ${{S}_{n}}=\dfrac{n}{2}[a+\{a+(n-1)d\}]$
Substituting the value of Sn and d, we get:
⇒ $-14=\dfrac{n}{2}[a+\{a+(n-1)2\}]$
⇒ -28 = n(2a + 2n - 2)
Substituting the value of a from equation (1) above, we will get:
⇒ -28 = n[2(6 - 2n) + 2n -2]
On multiplying to expand the inside brackets:
⇒ -28 = n(12 - 4n + 2n - 2)
⇒ -28 = n(10 - 2n)
Now, multiplying the RHS and bringing all the terms to LHS:
⇒ $2{{n}^{2}}-10n-28=0$
⇒ ${{n}^{2}}-5n-14=0$
On splitting the middle term and factorizing:
⇒ ${{n}^{2}}-7n+2n-14=0$
⇒ n(n - 7) + 2(n - 7) = 0
⇒ (n - 7)(n + 2) = 0
Since the product 0 is possible only if one of the multipliers is 0, we can say that:
⇒ n - 7 = 0 or n + 2 = 0
⇒ n = 7 or n = -2.
Since n is the number of terms, it cannot be negative. Therefore, n = 7.
And using equation (1) above:
a = 6 - 2n = 6 - 2 × 7 = 6 - 14 = -8.
Therefore, n = 7 and a = -8 is the answer.

Note: The question can also be solved easily by writing the Arithmetic Progression starting from the nth term itself. Since the common difference d = 2, we need to subtract 2 when we start from the nth term and go backwards. The sequence of numbers will be:
4, 4 - 2×1, 4 - 2×2, 4 - 2×3, 4 - 2×4, 4 - 2×5, 4 - 2×6, 4 - 2×7, 4 - 2×8, ... and so on.
= 4, 2, 0, -2, -4, -6, -8, -10, -12, ... and so on.
If we sum the above series, we find that 4 + 2 + 0 - 2 - 4 - 6 - 8 = -14.
Therefore, we need n = 7 terms to give the sum as -14 and the final term is a = -8, which answers the question.