Albarellak

2020-12-24

Given the first three terms of an arithmetic progression as $2x,\text{}x\text{}+\text{}4\text{}and\text{}2x\text{}-\text{}6$ respectively. Find the value of x and then give the first five terms of this arithmetic sequence.

Sadie Eaton

Skilled2020-12-25Added 104 answers

Step 1
let ${a}_{1},\text{}{a}_{2},\text{}{a}_{3},\text{}{a}_{4},\text{}{a}_{5}$ be the terms of A.P with common difference "d"
according to question
${a}_{1}=2x$

${a}_{2}=x\text{}+\text{}4$

${a}_{3}=2x\text{}-\text{}6$

$d={a}_{2}\text{}-\text{}{a}_{1}$

$d=x\text{}+\text{}4\text{}-\text{}2x$

$d=4\text{}-\text{}x$
also
$d={a}_{3}\text{}-\text{}{a}_{2}$

$d=2x\text{}-\text{}6\text{}-\text{}x\text{}-\text{}4$

$d=x\text{}-\text{}10$

$\text{as}\text{}x\text{}-\text{}10=4\text{}-\text{}x$

$2x=14$

$x=7$
Step 2
$\text{so as}\text{}{a}_{1}=2x$

$\Rightarrow \text{}{a}_{1}=14$

${a}_{2}={a}_{1}\text{}+\text{}d$

${a}_{2}=14\text{}-\text{}3$

${a}_{2}=11$

${a}_{3}={a}_{2}\text{}+\text{}d$

${a}_{3}=11\text{}-\text{}3$

${a}_{3}=8$

${a}_{4}={a}_{3}\text{}+\text{}d$

${a}_{4}=8\text{}-\text{}3$

${a}_{4}=5$

${a}_{5}=5\text{}-\text{}3$

${a}_{5}=2$

$\text{so}\text{}14,\text{}11,\text{}8,\text{}5,\text{}2$

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