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

Let A.P. be 3,3+d, 3+2 d, 3+3 dnSo, mathbf3+mathbfd=mathbfa, mathbf3+mathbf2 mathbfd=mathbfb, mathbf3+mathbf3 mathbfd=mathbfcnAlso, Let G.P. be 3,3 r, 2 r^2, 3 r^3nSo, 3 r=a-1,3 r^2=b+1,3 r^3=c+9nOn comparing,nn n 3 r=3+ d-1 => d=3 r-2 n 3 r^2=3+2 d+1 => 3 r^2=2 d+4n nn nSolve (1) and (2), we getnn r=2nnn a=7, b=11, c=15nnA.M. of a, b, c => (7+11+15 / 3)=11

Let 3,a,b,c be in A.P. and 3,a - 1,b + 1,c + 9 be in G... | Sequence & Series (Mathematics)

Let A.P. be

3,3+d, 3+2 d, 3+3 d

\nSo,

\mathbf{3}+\mathbf{d}=\mathbf{a}, \mathbf{3}+\mathbf{2} \mathbf{d}=\mathbf{b}, \mathbf{3}+\mathbf{3} \mathbf{d}=\mathbf{c}

\nAlso, Let G.P. be

3,3 \mathrm{r}, 2 \mathrm{r}^2, 3 \mathrm{r}^3

\nSo,

3 r=a-1,3 r^2=b+1,3 r^3=c+9

\nOn comparing,\n

\n\begin{aligned}\n& 3 \mathrm{r}=3+\mathrm{d}-1 \Rightarrow \mathrm{~d}=3 \mathrm{r}-2 \\\n& 3 \mathrm{r}^2=3+2 \mathrm{~d}+1 \Rightarrow 3 \mathrm{r}^2=2 \mathrm{~d}+4\n\end{aligned}\n

\n \nSolve (1) and (2), we get\n

\n\mathrm{r}=2\n

\n\mathrm{a}=7, \mathrm{~b}=11, \mathrm{c}=15\n

\nA.M. of

\mathrm{a}, \mathrm{b}, \mathrm{c} \Rightarrow \frac{7+11+15}{3}=11

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

Let a,b be two non-zero real numbers. If p and r are the roots of the equation x^{2} - 8ax + 2a = 0 and q and s are the roots of the equation \end{enumerate} x^{2} + 12bx + 6b = 0, such that \frac{1}{p},\frac{1}{q},\frac{1}{r},\frac{1}{s} are in A.P., then a^{- 1} - b^{- 1} is equal to

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

Number of 4 -digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11 , is equal to

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