Question

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.

Answer 1

Arithmetic Progression (A.P.)

Let the first term of the A.P. be \( a \), and the common difference be \( d \).

The nth term of an A.P. is given by:

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

Given:

• Third term is 16: \( a_3 = a + 2d = 16 \) — (1)
• Seventh term exceeds the fifth term by 12:

That is,

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

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

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

So, 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 formula for the sum of the first \( n \) terms is:

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

Substitute 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.