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Mathematics

How many terms of the series 18 + 15 + 12 + ……. when added together will give 45?

AP

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Answer

Let n terms be added to get sum of 45.

In above A.P., a = 18 and d = 15 - 18 = -3.

S=n2[2a+(n−1)d]⇒45=n2(2×18+(n−1)×(−3))⇒45=n2(36−3n+3)⇒45=n2(39−3n)⇒90=n(39−3n)⇒3n(13−n)=90⇒n(13−n)=30⇒13n−n2=30⇒n2−13n+30=0⇒n2−10n−3n+30=0⇒n(n−10)−3(n−10)=0⇒(n−3)(n−10)=0⇒n=3,10.S = \dfrac{n}{2}[2a + (n - 1)d] \\[1em] \Rightarrow 45 = \dfrac{n}{2}(2 \times 18 + (n - 1) \times (-3)) \\[1em] \Rightarrow 45 = \dfrac{n}{2}(36 - 3n + 3) \\[1em] \Rightarrow 45 = \dfrac{n}{2}(39 - 3n)\\[1em] \Rightarrow 90 = n(39 - 3n) \\[1em] \Rightarrow 3n(13 - n) = 90 \\[1em] \Rightarrow n(13 - n) = 30 \\[1em] \Rightarrow 13n - n^2 = 30 \\[1em] \Rightarrow n^2 - 13n + 30 = 0 \\[1em] \Rightarrow n^2 - 10n - 3n + 30 = 0 \\[1em] \Rightarrow n(n - 10) - 3(n - 10) = 0 \\[1em] \Rightarrow (n - 3)(n - 10) = 0 \\[1em] \Rightarrow n = 3, 10.

Hence, n = 3 or 10.

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