In an A.P., if $a = 8$ and $a_{10} = -19$, then value of $d$ is:

Question

Let $a_1,a_2,a_3,a_4$ be an A.P. of four terms such that each term of the A.P. and its common difference are integers. If $a_1+a_2+a_3+a_4=48$ and $a_1^2a_2a_3a_4+1^4=361$, then the largest term of the A.P. is equal to

Answer 1

The formula for the $n$-th term of an arithmetic progression (A.P.) is:

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

Where:

$a_n$ represents the $n$-th term,
$a$ is the first term,
$d$ is the common difference.

Given information:

First term, $a = 8$,
10th term, $a_{10} = -19$.
Objective: Find the common difference, $d$.

Applying the formula for the 10th term with the given values:

\[ a_{10} = a + (10 - 1)d \implies -19 = 8 + 9d \]

Solving for $d$:

\[ -19 - 8 = 9d \implies -27 = 9d \implies d = \frac{-27}{9} \implies d = -3 \]

The common difference is:

$d = -3$

In an A.P., if \(a = 8\) and \(a_{10} = -19\), then value of \(d\) is:

The Correct Option is D

Solution and Explanation

Top Questions on Arithmetic Progression

Questions Asked in CBSE Class X exam