Question

Find the number of terms of the AP 18, 15, 12,_ _ _ so that their sum is 45. Explain the double answer.
(a) \[n=3,10\]
(b) \[n=2,10\]
(c) \[n=N\]
(d) \[n=10\]

Answer 1

Hint: In this question, the first term of the given series will be the value of a and the difference between consecutive terms will be the common difference d. Now, we need to find the sum of the given arithmetic progression series using the formula \[{{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\] and then equate it to 45 which on further simplification gives the value of n.

Complete step by step answer:
ARITHMETIC PROGRESSION (A.P)
A sequence in which the difference of two consecutive terms is constant, is called Arithmetic progression.
Where, a is called the first term of the series and d is called the common difference of the series
Here, the value of the common difference is given by
\[d={{a}_{2}}-{{a}_{1}}\]'
As we already know that the sum of n terms of an arithmetic progression is given by the formula
\[{{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\]
Now, the given series in the question is 18, 15, 12,_ _ _
Now, we have the first term of this series as 18 which gives
\[a=18\]
Let us now find the common difference of this series
\[\Rightarrow d={{a}_{2}}-{{a}_{1}}\]
Now, on substituting the respective values we get,
\[\Rightarrow d=15-18\]
Now, on further simplification we get,
\[\therefore d=-3\]
Let us assume that the number of terms in the given series as n
Here, given that sum of the series is 45
Now, from the formula of sum of n terms of an AP we have
\[\Rightarrow {{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\]
Now, on substituting the respective values we get,
\[\Rightarrow 45=\dfrac{n}{2}\left( 2\times 18+\left( n-1 \right)\times \left( -3 \right) \right)\]
Now, this can be further written in the simplified form as
\[\Rightarrow 90=n\left( 36+3-3n \right)\]
Now, on simplifying further this can be written as
\[\Rightarrow 90=39n-3{{n}^{2}}\]
Now, on rearranging the terms we get,
\[\Rightarrow 3{{n}^{2}}-39n+90=0\]
Now, on taking the common term out and simplifying we get,
\[\Rightarrow {{n}^{2}}-13n+30=0\]
Now, this can be further written as
\[\Rightarrow {{n}^{2}}-3n-10n+30=0\]
Now, on simplifying this further we get,
\[\Rightarrow n\left( n-3 \right)-10\left( n-3 \right)=0\]
Now, on taking out the common terms we can further write it as
\[\Rightarrow \left( n-3 \right)\left( n-10 \right)=0\]
Now, on further simplification we get,
\[\therefore n=3,10\]
Thus, there are two values of n because as this series is a decreasing AP there will be negative terms as the number of terms increases so the same sum can occur twice.
Hence, the correct option is (a).

Note:
Instead of using the factorisation method to get the values of n we can also use the direct formula and then simplify further which also gives the same result.
It is important to note that as the equation in n is a quadratic we get 2 values of n possible so we need to consider both of them because the series is a decreasing AP which gives negative terms on increasing the n values which further decreases the sum.
It is also to be noted that while substituting the respective values and simplifying we should not neglect any of the terms or substitute incorrectly.