How many terms of the AP 18 16, 14, 12

Question:

How many terms of the AP 18 16, 14, 12, …. are needed to give the sum 78? Explain the double answer.

Solution:

To Find: Number of terms required to make the sum 78.

Here $a=18, d=-2$

Let n be the number of terms required to make the sum 78.

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$78=\frac{n}{2}[2 \times 18+(n-1)(-2)]$

$\Rightarrow 78 \times 2=36 n-2 n^{2}+2 n$

$\Rightarrow n^{2}-19 n+78=0$

$\Rightarrow n^{2}-6 n-13 n+78=0$

$\Rightarrow n(n-6)-13(n-6)=0$

$\Rightarrow(n-13)(n-6)=0$

Either n = 13 or n = 6

Explanation: Since the given $\mathrm{AP}$ is a decreasing progression where $\mathrm{a}_{\mathrm{n}-1}>\mathrm{a}_{\mathrm{n}}$, it is bound to have negative values in the series. $S_{n}$ is maximum for $n=9$ or $n=10$ since $T_{10}$ is $0\left(S_{10}=S_{9}=S_{\max }=90\right)$. The sum of 78 can be attained by either adding 6 terms or 13 terms so that negative terms from $T_{11}$ onward decrease the maximum sum to 78 .

 

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