Find the A.P. whose third term is 16 and seventh term exceeds the fifth term by 12. Also, find the sum of first 29 terms of the A.P.

Question

In which of the following situations, does the list of numbers involved make as arithmetic progression and why?
(1) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.
(2) The amount of air present in a cylinder when a vacuum pump removes \(\frac{1}{4}\) of the air remaining in the cylinder at a time.
(3) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.
(4) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.

Answer 1

Arithmetic Progression (A.P.)

An arithmetic progression (A.P.) is defined by its first term \( a \) and common difference \( d \).

The nth term of an A.P. is:

\[ a_n = a + (n - 1)d \]

Given:

The third term is 16: \( a_3 = a + 2d = 16 \) — (1)
The seventh term exceeds the fifth term by 12:

This means,

\[ a_7 = a_5 + 12 \Rightarrow a + 6d = a + 4d + 12 \Rightarrow 6d = 4d + 12 \Rightarrow 2d = 12 \Rightarrow d = 6 \]

Substitute \( d = 6 \) into equation (1):

\[ a + 2(6) = 16 \Rightarrow a + 12 = 16 \Rightarrow a = 4 \]

Therefore, the first term is \( a = 4 \) and the common difference is \( d = 6 \).

The A.P. is:

\[ 4,\ 10,\ 16,\ 22,\ \ldots \]

Sum of the First 29 Terms \( S_{29} \):

The sum of the first \( n \) terms is:

\[ S_n = \frac{n}{2}[2a + (n - 1)d] \]

Substituting the values:

\[ S_{29} = \frac{29}{2}[2(4) + (29 - 1)(6)] = \frac{29}{2}[8 + 168] = \frac{29}{2}[176] = 29 \times 88 \]

\[ 29 \times 88 = 2552 \Rightarrow \boxed{S_{29} = 2552} \]

Conclusion:

The arithmetic progression is \( 4, 10, 16, \ldots \), and the sum of the first 29 terms is 2552.