If the mean of x, x+3,x+6,x+9,x+12,x+15.is 18 find the mean of first f

1.	If the mean of x, x+3,x+6,x+9,x+12,x+15.is 18 find the mean of first four observation
Answer» Given : Given observations = X,(x + 3),(x + 6),(x + 9),(x + 12),(x + 15). Mean of all the observations = 18. No. if term = 6 To find : Mean of first four observations. Solution : We KNOW the FORMULA for mean,i.e, $\boxed{\bf{Mean = \dfrac{\Sigma x_{i}}{n}}}$ Where : $\bf{\Sigma x_{i}}$ = Sum of all observations. $\bf{n}$ = No. of all observations. By using the formula for mean and substituting the values in it, we get : $:\implies \bf{Mean = \dfrac{\Sigma x_{i}}{n}} \\ \\ \\$ $:\implies \bf{Mean = \dfrac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6}}{n}} \\ \\ \\$ $:\implies \bf{18 = \dfrac{x + (x + 3) + (x + 6) + (x + 9) + (x + 12) + (x + 15)}{6}} \\ \\ \\$ $:\implies \bf{18 = \dfrac{(x + x + x + x + x + x) + (3 + 6 + 9 + 12 + 15)}{6}} \\ \\ \\$ $:\implies \bf{18 = \dfrac{6x + 45}{6}} \\ \\ \\$ $:\implies \bf{18 \times 6 = 6x + 45} \\ \\ \\$ $:\implies \bf{108 = 6x + 45} \\ \\ \\$ $:\implies \bf{108 - 45 = 6x} \\ \\ \\$ $:\implies \bf{63 = 6x} \\ \\ \\$ $:\implies \bf{\dfrac{63}{6} = x} \\ \\ \\$ $:\implies \bf{10.5 = x} \\ \\ \\$ $\boxed{\therefore \bf{x = 10.5}} \\ \\ \\$ Hence the value of x 10.5. From the above equation , we get : First term = x = 10.5 Second term = (x + 3) = 13.5 Third term = (x + 6) = 16.5 Fourth term = (x + 9) = 19.5 Fifth term = (x + 12) = 22.5 Sixth term = (x + 15) = 25.5 Now to find the mean of first four terms of the observation. $\boxed{\bf{Mean = \dfrac{\Sigma x_{i}}{n}}}$ Where : $\bf{\Sigma x_{i}}$ = Sum of all observations. $\bf{n}$ = No. of all observations. By using the formula for mean and substituting the values in it, we get : $:\implies \bf{Mean = \dfrac{\Sigma x_{i}}{n}} \\ \\ \\$ $:\implies \bf{Mean = \dfrac{x_{1} + x_{2} + x_{3} + x_{4}}{n}} \\ \\ \\$ $:\implies \bf{Mean = \dfrac{10.5 + 13.5 + 16.5 + 19.5}{4}} \\ \\ \\$ $:\implies \bf{Mean = \dfrac{10.5 + 13.5 + 16.5 + 19.5}{4}} \\ \\ \\$ $:\implies \bf{Mean = \dfrac{60}{4}} \\ \\ \\$ $:\implies \bf{Mean = 15} \\ \\ \\$ $\boxed{\therefore \bf{Mean = 15}} \\ \\ \\$ Hence the mean of first four terms of the observation is 15.