The sum of the first 5 terms of an AP is 30 and the sum of the four term from T6 to T9 (inclusive) is 69. Find the AP

The AP = 1, ⁷/₂, 6, ¹⁷/₂, 11

Step-by-step explanation:

From the first statement,

S₅ = ⁵/₂(2a + ( n - 1 )d }    = 30

       5(2a  + 4d )d           = 60

       10a    + 20d             = 60

reduce to lowest term to easy calculation by dividing through by 10

                      a + 2d       = 6 1

second statement

sum of the next 4 terms inclusive

T₉  = ⁹/₂(2a + 8d )   = 69

        9(2a  + 8d )    = 30 + 69

            18a + 72d   = 99 x 2

            18a + 72d   = 198

divide through by 18 to reduce to lowest time

                a  + 4d    = 11 2

Now solve the two equation simultaneously to find a and d

                 a + 2d = 6

                 a + 4d = 11

                     -2d  = -5

                          d = ⁵/₂.

Now substitute  for d to get a

               a  + 2(⁵/₂) = 6

                a + 5       = 6

                           a   = 6 - 5

                            a = 1.

Therefore the AP = 1 , ⁷/₂ , 6 , ¹⁷/₂ , 11 ,

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an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.