Question

An AP consists of 55 terms of which 5th term is 14 and last term is 144. Find the 29th term

Answer 1

Answer:

29th term = 76.4

Step-by-step explanation:

Total terms = 55

aₙ = a + (n - 1) d

5th term = a₅ = 14

   a + (5 - 1)*d = 14

   a + 4d = 14  --------------------(I)

last term = a₅₅ = 144

               a + (55 - 1)d = 144

      a + 54d = 144----------------(II)

Multiply equation (II) by (-1)

(I)             a +  4d = 14

(II)*(-1)     -a - 54d = -144  {Now add and thus a will be eliminated}

                   -50d = -130

 d = -130/-50

d = 2.6

Plug in d = 2.6 in equation (I)

a + 4* 2.6 = 14

  a  + 10.4 = 14

              a = 14 - 10.4

              a = 3.6

29th term = a₂₉ =  3.6 + (29 - 1)* 2.6

                          = 3.6 + 28* 2.6

                          = 3.6 + 72.8

                     a₂₉ =  76.4

Answer 2

Answer:

76.4

Step-by-step explanation:

Let the first term and common difference of the AP be a and D respectively.

$t_5 = 14 \implies \: a + 4d = 14....(1) \\ t_{55} = 144 \implies \: a + 54d = 144....(2) \\ subtract \: eqution \: (1) \: from \: equatin \: (2) \\ \\ a + 54d - (a + 4d) = 144 - 14 \\ a + 54d - a - 4d = 130 \\ 50d = 130 \\ d = \frac{130}{50} \\ \huge \red{ \boxed{ d = 2.6}} \\ substituting \: d = 2.6 \: in \: equation \: (1) \\ \\ a + 4 \times 2.6 = 14 \\ a + 10.4 = 14 \\ a = 14 - 10.4 \\ \huge \purple{ \boxed{ a = 3.6}} \\ \\ t_{29} = a + 28d \\ t_{29} = 3.6 + 28 \times 2.6 \\ t_{29} = 3.6 + 72.8 \\\huge \orange{ \boxed{ t_{29} = 76.4}}$