Question

In an A.P., \(15^{th}\) term exceeds the \(8^{th}\) term by 21. If sum of first 10 terms is 55, then form the A.P.

Hint:

The difference between any two terms \(a_{p}\) and \(a_{q}\) is simply \((p - q)d\). Here, \(a_{15} - a_{8} = 7d\), which allows you to find \(d\) instantly.

Answer 1

Let the A.P. have first term \(a\) and common difference \(d\).

Given 1:
The 15th term exceeds the 8th term by 21.
\[ a_{15} - a_8 = 21 \] Use \(a_n = a + (n-1)d\):
\[ [a + 14d] - [a + 7d] = 21 \] \[ 7d = 21 \] \[ d = 3 \]

Given 2:
Sum of first 10 terms is 55.
\[ S_{10} = 55 \] Formula: \[ S_n = \frac{n}{2}[2a + (n-1)d] \] Substitute: \[ 55 = \frac{10}{2}[2a + 9 \cdot 3] \] \[ 55 = 5(2a + 27) \] Divide both sides by 5:
\[ 11 = 2a + 27 \] \[ 2a = -16 \] \[ a = -8 \]

Thus, the A.P. is:
First term = -8
Common difference = 3

A.P. = \[ -8,\ -5,\ -2,\ 1,\ 4,\ 7,\ \ldots \]
Final Answer:
The required A.P. is \[ \boxed{-8,\ -5,\ -2,\ 1,\ 4,\ 7,\ \ldots} \]