Question

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

Answer 1

76.4

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 = (130)/(50) \\ \huge \red{ \boxed{ d = 2.6}} \\ substituting \: d = 2.6 \: in \: equation \: (1) \\ \\ a + 4 * 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 * 2.6 \\ t_(29) = 3.6 + 72.8 \\\huge \orange{ \boxed{ t_(29) = 76.4}} \\`

Answer 2

29th term = 76.4

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