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How many terms of the arithmetic progression $45, 39, 33, ........$ must be taken so that their sum is $180$? Explain the double answer.
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$45, 39, 33, .......$
a = $45$, d = $- 6$
Sn = $180$
$180 = \frac{n}{2} [2 \times 45 + (n - 1) (-6)]$
$180 = \frac{n}{2} [90-6n + 6]$
$\Rightarrow 360 = 96n - 6n^2$
$\Rightarrow 6n^2 - 96n + 360 = 0$
$\Rightarrow n^2 - 16n + 60 = 0 \Rightarrow (n - 10) (n - 6) = 0$
n - 10 = $0$, n - 6 = $0 \Rightarrow n = 10, 6$
We get two values of 'n' as sum of $7^{th}$ term to $10^{th}$ term is zero as some terms are negative and some are positive.
a = $45$, d = $- 6$
Sn = $180$
$180 = \frac{n}{2} [2 \times 45 + (n - 1) (-6)]$
$180 = \frac{n}{2} [90-6n + 6]$
$\Rightarrow 360 = 96n - 6n^2$
$\Rightarrow 6n^2 - 96n + 360 = 0$
$\Rightarrow n^2 - 16n + 60 = 0 \Rightarrow (n - 10) (n - 6) = 0$
n - 10 = $0$, n - 6 = $0 \Rightarrow n = 10, 6$
We get two values of 'n' as sum of $7^{th}$ term to $10^{th}$ term is zero as some terms are negative and some are positive.