Question 60Write the first five terms of the sequence given by (3)n(\sqrt{3})^n(3)n, n ∈ N.(a) Is the sequence an A.P. or G.P.?(b) If the sum of its first ten terms is p(3+3)p(3 + \sqrt{3})p(3+3), find the value of p.

Question

Question 60Write the first five terms of the sequence given by (3)n(\sqrt{3})^n(3​)n, n ∈ N.(a) Is the sequence an A.P. or G.P.?(b) If the sum of its first ten terms is p(3+3)p(3 + \sqrt{3})p(3+3​), find the value of p.

Accepted Answer

Terms of the sequence given by (3)n(\sqrt{3})^n(3​)n are :⇒ (3)1,(3)2,(3)3,(3)4,(3)5(\sqrt{3})^1, (\sqrt{3})^2, (\sqrt{3})^3, (\sqrt{3})^4, (\sqrt{3})^5(3​)1,(3​)2,(3​)3,(3​)4,(3​)5, .......⇒ 3,3,33,9,93\sqrt{3}, 3, 3\sqrt{3}, 9, 9\sqrt{3}3​,3,33​,9,93​(a) Ratio between terms = 33=3\dfrac{3}{\sqrt{3}} = \sqrt{3}3​3​=3​.Hence, the sequence is a G.P. with common ratio = 3\sqrt{3}3​.(b) By formula,Sum of G.P. (S) = a(rn−1)(r−1)\dfrac{a(r^n - 1)}{(r - 1)}(r−1)a(rn−1)​Substituting values we get :⇒S10=3[(3)10−1]3−1⇒p(3+3)=3(243−1)3−1⇒p=3(243−1)(3−1)(3+3)⇒p=3×242(3−1)3(3+1)⇒p=2423−1⇒p=2422=121.\Rightarrow S_{10} = \dfrac{\sqrt{3}[(\sqrt{3})^{10} - 1]}{\sqrt{3} - 1} \[1em] \Rightarrow p(3 + \sqrt{3})= \dfrac{\sqrt{3}(243 - 1)}{\sqrt{3} - 1} \[1em] \Rightarrow p = \dfrac{\sqrt{3}(243 - 1)}{(\sqrt{3} - 1)(3 + \sqrt{3})} \[1em] \Rightarrow p = \dfrac{\sqrt{3} \times 242}{(\sqrt{3} - 1)\sqrt{3}(\sqrt{3} + 1)} \[1em] \Rightarrow p = \dfrac{242}{3 - 1} \[1em] \Rightarrow p = \dfrac{242}{2} = 121.⇒S10​=3​−13​[(3​)10−1]​⇒p(3+3​)=3​−13​(243−1)​⇒p=(3​−1)(3+3​)3​(243−1)​⇒p=(3​−1)3​(3​+1)3​×242​⇒p=3−1242​⇒p=2242​=121.Hence, p = 121.

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Question 60