Question
Write the value of $x$ for which $2x, x + 10$ and $3x + 2$ are in A.P.
✓
Answer
Here, we are given three terms, First term $\left(a_1\right)=2 x$
Second term $\left(a_2\right)=x+10$ Third term $\left(a_3\right)=3 x+2$
We need to find the value of $x$ for which these terms are in A.P.
So, in an A.P. the difference of two adjacent terms is always constant.
So, we get, $d=a_2-a_1 d=(x+10)-(2 x) d=x+10-2 x d=10-x$.....(i)
Also, $d=a 3-a 2 d=(3 x+2)-(x+10)$
$d=2 x-8 \ldots . . . \text { (ii) }$
Now, on equatin (i) and (ii),
we get, $10-x=2 x-82 x+x=10+83 x=18$
$x=\frac{18}{3} x=6$ Therefore, for $x=6$, these three terms will from an A.P.
Second term $\left(a_2\right)=x+10$ Third term $\left(a_3\right)=3 x+2$
We need to find the value of $x$ for which these terms are in A.P.
So, in an A.P. the difference of two adjacent terms is always constant.
So, we get, $d=a_2-a_1 d=(x+10)-(2 x) d=x+10-2 x d=10-x$.....(i)
Also, $d=a 3-a 2 d=(3 x+2)-(x+10)$
$d=2 x-8 \ldots . . . \text { (ii) }$
Now, on equatin (i) and (ii),
we get, $10-x=2 x-82 x+x=10+83 x=18$
$x=\frac{18}{3} x=6$ Therefore, for $x=6$, these three terms will from an A.P.
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