If \ a_i \_i = 1^n, where n is an evenn integer, is an arithmetic progression with common difference 1 , andn sum _i = 1^nmspace2mu a_i = 192,sum _i = 1^n/2mspace2mu a_2i = 120,n and then n is equal to

(b) : Here, d=1, a_1, a_2, a_3 ... a_n in A.P.nn nsum _i=1^n a_i =192 => (n / 2)[2 a_1+(n-1) 1]=192 n => n a_1+(n(n-1) / 2)=192 => n a_1=192-(n(n-1) / 2) \ (i)n nn nAlso, sum _i=1^n / 2 a_2 i=120 => a_2+a_4+ * s+a_n=120nn => a_1+1+a_1+3+ * s .+a_1+(n-1) * 1=120nn nn n => (n / 2)(a_1)+[1+3+ * s * +(n-1)]=120 n => (n / 2)(a_1)+((n / 2))^2=120n nn[Sum of first n odd natural numbers =n^2 ]nn => (1 / 2)(192-(n(n-1) / 2))+(n^2 / 4)=120nn[Using (i)]nn => 96+(n / 4)=120 => n=96n

If \{ a_{i} \}_{i = 1}^{n}, where n is an even integ... | Sequence & Series (Mathematics)

(b) : Here,

d=1, a_1, a_2, a_3 \ldots a_n \in

A.P.\n

\n\begin{aligned}\n\sum_{i=1}^n a_i & =192 \Rightarrow \frac{n}{2}\left[2 a_1+(n-1) 1\right]=192 \\\n& \Rightarrow n a_1+\frac{n(n-1)}{2}=192 \Rightarrow n a_1=192-\frac{n(n-1)}{2} \#(i)\n\end{aligned}\n

\n \nAlso,

\sum_{i=1}^{n / 2} a_{2 i}=120 \Rightarrow a_2+a_4+\cdots+a_n=120

\n\Rightarrow a_1+1+a_1+3+\cdots .+a_1+(n-1) \cdot 1=120\n

\n \n

\n\begin{aligned}\n& \Rightarrow \frac{n}{2}\left(a_1\right)+[1+3+\cdots \cdot+(n-1)]=120 \\\n& \Rightarrow \frac{n}{2}\left(a_1\right)+\left(\frac{n}{2}\right)^2=120\n\end{aligned}\n

\n[Sum of first

n

odd natural numbers

=n^2

]\n

\n\Rightarrow \frac{1}{2}\left(192-\frac{n(n-1)}{2}\right)+\frac{n^2}{4}=120\n

\n[Using (i)]\n

\n\Rightarrow 96+\frac{n}{4}=120 \Rightarrow n=96\n

Explanation

Let a_{1},a_{2},,a_{n} be a given A.P. whose common difference is an integer and S_{n} = a_{1} + a_{2} + + a_{n}. If a_{1} = 1,a_{n} = 300 and 15 \leq n \leq 50, then the ordered pair ( S_{n - 4},a_{n - 4} ) is equal to

Let a_{1},a_{2},a_{3},. be a G.P. of increasing positive numbers. Let the sum of its 6^{{th}{~}} and 8^{{th}{~}} terms be 2 and the product of its 3^{{rd}{~}} and 5^{{th}{~}} terms be 1/9. Then 6( a_{2} + a_{4} )( a_{4} + a_{6} ) is equal to

For any odd integer n ? 1,n^{3} - (n - 1)^{3} + + ( - 1)^{n - 1}1^{3} =

The sum of the first n terms of the series (1^{2} + 2.2^{2} + 3^{2} + 2.4^{2} + 5^{2} + 2 . 6^{2} + ..) is (\frac{n(n + 1)^{2}}{2}, when n is even. when n is odd, the sum is

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If a_{1},a_{2},a_{3}, are in A.P. such that a_{1} + a_{7} + a_{16} = 40, then the sum of the first 15 terms of this A.P. is

Let \frac{1}{x_{1}},\frac{1}{x_{2}},,\frac{1}{x_{n}}( x_{i} eq 0 .\ for . \ i = 1,2,,n ) be in A.P. such that x_{1} = 4 and x_{21} = 20. If n is the least positive integer for which x_{n} > 50, then \sum_{i = 1}^{n}\mspace{2mu}( \frac{1}{x_{i}} ) is equal to

Let 3,a,b,c be in A.P. and 3,a - 1,b + 1,c + 9 be in G.P. Then, the arithmetic mean of a,b and c is

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